{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0001","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0ffdfd5737e2f0622f47ed6921f41e5cadfd43a9c3d3ce2c2d59ee9dfe1381a","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133001,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[1,1],[2,1],[3,1]],\"index\":0}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[1, 2, 3] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=0. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([1, 2, 3], a_n=0_after_the_list, 0, 1, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([1, 2, 3], a_n=0_after_the_list, 0, 1, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0002","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff09ca00044e634c925cb345fb5e06f315bc1cc8003f3109b7b4b37723406294","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133002,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[2,1],[4,1],[6,1],[8,1]],\"index\":1}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[2, 4, 6, 8] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=1. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([2, 4, 6, 8], a_n=0_after_the_list, 1, 4, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([2, 4, 6, 8], a_n=0_after_the_list, 1, 4, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0003","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6944ec6657ec600c2c2d8b5adc83e82be2a93471435e37a07a078874f88e2e45","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133003,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[3,1],[6,1],[9,1],[12,1],[15,1]],\"index\":2}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[3, 6, 9, 12, 15] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=2. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([3, 6, 9, 12, 15], a_n=0_after_the_list, 2, 9, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([3, 6, 9, 12, 15], a_n=0_after_the_list, 2, 9, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0004","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c529d93f24080eeda406dc317b183741de2fe0d0b2101ec5544f02ecd5620f30","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133004,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[4,1],[8,1],[12,1],[16,1],[20,1],[24,1]],\"index\":3}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[4, 8, 12, 16, 20, 24] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=3. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([4, 8, 12, 16, 20, 24], a_n=0_after_the_list, 3, 16, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([4, 8, 12, 16, 20, 24], a_n=0_after_the_list, 3, 16, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0005","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3a0bbf8aaaca53395bab563e5d6ad0efc8896660eafef526d010e77f4efd3d5","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133005,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[5,1],[10,1],[15,1]],\"index\":4}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[5, 10, 15] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=4. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([5, 10, 15], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([5, 10, 15], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0006","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24b3132686220cb7a34fe37ad109e2571eb79dc57aad63defaddcbd94386093d","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133006,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[6,1],[12,1],[18,1],[24,1]],\"index\":5}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[6, 12, 18, 24] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=5. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([6, 12, 18, 24], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([6, 12, 18, 24], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0007","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d732e1fd8db2ceac34c9d87160368f0ee08316a47eaae7e4a3908580f9f67196","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133007,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[7,1],[14,1],[21,1],[28,1],[35,1]],\"index\":6}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[7, 14, 21, 28, 35] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=6. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([7, 14, 21, 28, 35], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([7, 14, 21, 28, 35], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0008","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"212e9362bf26ac4444980e1f5a88c7c89ada13f38af8e2cf41ebc4b6e589180a","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133008,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[8,1],[16,1],[24,1],[32,1],[40,1],[48,1]],\"index\":7}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[8, 16, 24, 32, 40, 48] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=7. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([8, 16, 24, 32, 40, 48], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([8, 16, 24, 32, 40, 48], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0009","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6ec7b39d488433cfbd07b0caa7d7c974608954920754c74052def29e25e9fb3","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133009,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[9,1],[18,1],[27,1]],\"index\":8}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[9, 18, 27] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=8. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([9, 18, 27], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([9, 18, 27], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0010","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e285f9f7f7640980274bd67973c1faece68cc89c0699b14753cfe78e967c5e37","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133010,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[10,1],[20,1],[30,1],[40,1]],\"index\":0}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[10, 20, 30, 40] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=0. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([10, 20, 30, 40], a_n=0_after_the_list, 0, 10, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([10, 20, 30, 40], a_n=0_after_the_list, 0, 10, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0011","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ecd6a69976956f219f7b301bad6af5361f1b2c124e597cd59bae6dd485e05f0","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133011,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[11,1],[22,1],[33,1],[44,1],[55,1]],\"index\":1}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[11, 22, 33, 44, 55] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=1. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([11, 22, 33, 44, 55], a_n=0_after_the_list, 1, 22, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([11, 22, 33, 44, 55], a_n=0_after_the_list, 1, 22, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0012","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45da7daf23cc06a54a0a359f3a916b9efd29161476f5d5be11342cbe328c6d0e","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133012,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[12,1],[24,1],[36,1],[48,1],[60,1],[72,1]],\"index\":2}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[12, 24, 36, 48, 60, 72] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=2. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([12, 24, 36, 48, 60, 72], a_n=0_after_the_list, 2, 36, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([12, 24, 36, 48, 60, 72], a_n=0_after_the_list, 2, 36, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0013","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6528577f1e64312131390ee0993a907a04b62ea65ae4e51df142bdaf406a863f","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133013,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[13,1],[26,1],[39,1]],\"index\":3}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[13, 26, 39] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=3. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([13, 26, 39], a_n=0_after_the_list, 3, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([13, 26, 39], a_n=0_after_the_list, 3, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0014","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29dca8ce4d9fd100b15e60a785435b87df7564713ee93b67c939d4f4bfa471c0","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133014,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[14,1],[28,1],[42,1],[56,1]],\"index\":4}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[14, 28, 42, 56] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=4. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([14, 28, 42, 56], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([14, 28, 42, 56], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0015","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ccb479884526b7219771e225099655b4730a92067bb9c45795e681a95ffac567","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133015,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[15,1],[30,1],[45,1],[60,1],[75,1]],\"index\":5}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[15, 30, 45, 60, 75] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=5. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([15, 30, 45, 60, 75], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([15, 30, 45, 60, 75], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0016","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ba9c1e2002f2a66f80ad52862e55e44b6eb79f5e38d4547c0e8cf169767578d","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133016,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[16,1],[32,1],[48,1],[64,1],[80,1],[96,1]],\"index\":6}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[16, 32, 48, 64, 80, 96] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=6. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([16, 32, 48, 64, 80, 96], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([16, 32, 48, 64, 80, 96], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0017","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b43738f92c89230a15c2dafaaaf5827ffa36b6f95b54712d43f285ea3e3f999","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133017,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[17,1],[34,1],[51,1]],\"index\":7}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[17, 34, 51] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=7. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([17, 34, 51], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([17, 34, 51], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0018","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1bd93026394e8779824b232e7fcdc83fce30d2753b31afae94b814d14b7db464","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133018,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[18,1],[36,1],[54,1],[72,1]],\"index\":8}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[18, 36, 54, 72] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=8. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([18, 36, 54, 72], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([18, 36, 54, 72], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0019","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ea30249d501c3c763cb491d9ec1ee7930a178576b7ce9988b45f6262fb6832f","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133019,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[19,1],[38,1],[57,1],[76,1],[95,1]],\"index\":0}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[19, 38, 57, 76, 95] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=0. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([19, 38, 57, 76, 95], a_n=0_after_the_list, 0, 19, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([19, 38, 57, 76, 95], a_n=0_after_the_list, 0, 19, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0020","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0783b56c749ba603e5c8f0cecdf1fae7bb0f9c0ae65820e9ae69eb9aee089ce5","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133020,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[20,1],[40,1],[60,1],[80,1],[100,1],[120,1]],\"index\":1}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[20, 40, 60, 80, 100, 120] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=1. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([20, 40, 60, 80, 100, 120], a_n=0_after_the_list, 1, 40, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([20, 40, 60, 80, 100, 120], a_n=0_after_the_list, 1, 40, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0021","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6fb9cdb953cd879509d9d4b3e409fffde6f6088669c1ef796f84d1dd89292948","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133021,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[21,1],[42,1],[63,1]],\"index\":2}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[21, 42, 63] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=2. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([21, 42, 63], a_n=0_after_the_list, 2, 63, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([21, 42, 63], a_n=0_after_the_list, 2, 63, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0022","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dae893ed8b9e9a094105762be5ccbe788c3ecf7051a9b682225b5a662f101b3","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133022,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[22,1],[44,1],[66,1],[88,1]],\"index\":3}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[22, 44, 66, 88] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=3. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([22, 44, 66, 88], a_n=0_after_the_list, 3, 88, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([22, 44, 66, 88], a_n=0_after_the_list, 3, 88, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0023","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95c5c1f6946be49ee794ce30822a2d0b1a05e90fd7bdd02eb5e31d1173bc48c5","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133023,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[23,1],[46,1],[69,1],[92,1],[115,1]],\"index\":4}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[23, 46, 69, 92, 115] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=4. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([23, 46, 69, 92, 115], a_n=0_after_the_list, 4, 115, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([23, 46, 69, 92, 115], a_n=0_after_the_list, 4, 115, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0024","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ecec0a2049924c9a8f1dcb51b93887754ab998849acded3f39cc3ffbcf0534c4","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133024,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[24,1],[48,1],[72,1],[96,1],[120,1],[144,1]],\"index\":5}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[24, 48, 72, 96, 120, 144] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=5. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([24, 48, 72, 96, 120, 144], a_n=0_after_the_list, 5, 144, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([24, 48, 72, 96, 120, 144], a_n=0_after_the_list, 5, 144, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0025","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8af80a3ee26863d9fb9643abccee5ae8c8c20fbfb756f5ed55be1269ec8b399","family_id":"qf-mod44-ogf-coefficient-extraction","seed":202611133025,"semantic_template":"qf-mod44-ogf-coefficient-extraction/{\"finite_support_coefficients\":[[25,1],[50,1],[75,1]],\"index\":6}"},"fields":{"title":"Coefficient Extraction Uses the Declared Complete Zero Tail","stem":"Use finite_support_coefficients=[25, 50, 75] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Extract index=6. Give (finite_support_coefficients, zero_tail_law, index, coefficient, certificate).","answer":{"kind":"expression","value":"([25, 50, 75], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)","explanation":"Return the indexed coefficient or0 beyond support. Ordinary generating coefficients are a_n, not a_n/n! of an exponential generating function. A finite list with an explicit zero tail differs from an unknown continuation. The exact result, retaining the requested order, is ([25, 50, 75], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)."},"skills":["Coefficient Extraction Uses the Declared Complete Zero Tail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0026","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba287f9f33873ee14e1f8d17e3bb07b3c26a3546ec5f56804fd024ee2ff5d8a1","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133026,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[1,1],[2,1],[3,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[1, 2, 3] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([1, 2, 3], 1/2, 2, [[0, 1, 1], [1, 2, 1], [2, 3, 3/4]], 11/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([1, 2, 3], 1/2, 2, [[0, 1, 1], [1, 2, 1], [2, 3, 3/4]], 11/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0027","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bec576d0d739db954e24fc2f849bcb8bcb01d5284b4907c79a2a07f3ea2d0a4a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133027,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[2,1],[4,1],[6,1],[8,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[2, 4, 6, 8] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([2, 4, 6, 8], 1/3, 2, [[0, 2, 2], [1, 4, 4/3], [2, 6, 2/3]], 4, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([2, 4, 6, 8], 1/3, 2, [[0, 2, 2], [1, 4, 4/3], [2, 6, 2/3]], 4, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0028","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5fe66420b298251c353e18b39472f1887ad87779923446b8857c3156deb559a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133028,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[3,1],[6,1],[9,1],[12,1],[15,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[3, 6, 9, 12, 15] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([3, 6, 9, 12, 15], 1/4, 4, [[0, 3, 3], [1, 6, 3/2], [2, 9, 9/16], [3, 12, 3/16], [4, 15, 15/256]], 1359/256, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([3, 6, 9, 12, 15], 1/4, 4, [[0, 3, 3], [1, 6, 3/2], [2, 9, 9/16], [3, 12, 3/16], [4, 15, 15/256]], 1359/256, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0029","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9adf2f1d727155e597b8eef9b34eefcfff71785c90bb3752206e17aa5c70886b","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133029,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[4,1],[8,1],[12,1],[16,1],[20,1],[24,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[4, 8, 12, 16, 20, 24] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([4, 8, 12, 16, 20, 24], 1/2, 4, [[0, 4, 4], [1, 8, 4], [2, 12, 3], [3, 16, 2], [4, 20, 5/4]], 57/4, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([4, 8, 12, 16, 20, 24], 1/2, 4, [[0, 4, 4], [1, 8, 4], [2, 12, 3], [3, 16, 2], [4, 20, 5/4]], 57/4, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0030","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28bcad1384ecb2140be2664cb81bc1b032e747a910bbd71fbff64ac8933cdc9a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133030,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[5,1],[10,1],[15,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[5, 10, 15] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([5, 10, 15], 1/3, 2, [[0, 5, 5], [1, 10, 10/3], [2, 15, 5/3]], 10, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([5, 10, 15], 1/3, 2, [[0, 5, 5], [1, 10, 10/3], [2, 15, 5/3]], 10, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0031","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"838b5a3fb944aa92038e243aaa055b67dc6da7d139f661f1a6738533ffe60d26","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133031,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[6,1],[12,1],[18,1],[24,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[6, 12, 18, 24] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([6, 12, 18, 24], 1/4, 2, [[0, 6, 6], [1, 12, 3], [2, 18, 9/8]], 81/8, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([6, 12, 18, 24], 1/4, 2, [[0, 6, 6], [1, 12, 3], [2, 18, 9/8]], 81/8, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0032","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"536a34833c90a8dbb90661a9032ac88fcb3a1e63d1cbb72764b56f6d8afe4705","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133032,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[7,1],[14,1],[21,1],[28,1],[35,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[7, 14, 21, 28, 35] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([7, 14, 21, 28, 35], 1/2, 4, [[0, 7, 7], [1, 14, 7], [2, 21, 21/4], [3, 28, 7/2], [4, 35, 35/16]], 399/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([7, 14, 21, 28, 35], 1/2, 4, [[0, 7, 7], [1, 14, 7], [2, 21, 21/4], [3, 28, 7/2], [4, 35, 35/16]], 399/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0033","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4ddbd0ee97d58e62efcccd36b5fdc145c6b1a8b173ae4b294132ddecc771a04","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133033,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[8,1],[16,1],[24,1],[32,1],[40,1],[48,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[8, 16, 24, 32, 40, 48] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([8, 16, 24, 32, 40, 48], 1/3, 4, [[0, 8, 8], [1, 16, 16/3], [2, 24, 8/3], [3, 32, 32/27], [4, 40, 40/81]], 1432/81, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([8, 16, 24, 32, 40, 48], 1/3, 4, [[0, 8, 8], [1, 16, 16/3], [2, 24, 8/3], [3, 32, 32/27], [4, 40, 40/81]], 1432/81, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0034","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb5634a368af543bedfb7c6ef8da82b8958805924676893f275e98e37b36667a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133034,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[9,1],[18,1],[27,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[9, 18, 27] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([9, 18, 27], 1/4, 2, [[0, 9, 9], [1, 18, 9/2], [2, 27, 27/16]], 243/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([9, 18, 27], 1/4, 2, [[0, 9, 9], [1, 18, 9/2], [2, 27, 27/16]], 243/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0035","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dcf9cee1262bd38cc31b5fe58194d75c76af24d53b7e8e4fb246dcb22138daf","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133035,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[10,1],[20,1],[30,1],[40,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[10, 20, 30, 40] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([10, 20, 30, 40], 1/2, 2, [[0, 10, 10], [1, 20, 10], [2, 30, 15/2]], 55/2, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([10, 20, 30, 40], 1/2, 2, [[0, 10, 10], [1, 20, 10], [2, 30, 15/2]], 55/2, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0036","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b3a0c6dc75f129bd84b37dc3de0d1ea4e6ba30d998dc3b4b2573da73c9a50f0","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133036,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[11,1],[22,1],[33,1],[44,1],[55,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[11, 22, 33, 44, 55] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([11, 22, 33, 44, 55], 1/3, 4, [[0, 11, 11], [1, 22, 22/3], [2, 33, 11/3], [3, 44, 44/27], [4, 55, 55/81]], 1969/81, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([11, 22, 33, 44, 55], 1/3, 4, [[0, 11, 11], [1, 22, 22/3], [2, 33, 11/3], [3, 44, 44/27], [4, 55, 55/81]], 1969/81, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0037","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a2176467225497a655c2e69b5afb15258a5c1a669d4f0ca67cc8e3b1c63fd50","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133037,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[12,1],[24,1],[36,1],[48,1],[60,1],[72,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[12, 24, 36, 48, 60, 72] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([12, 24, 36, 48, 60, 72], 1/4, 4, [[0, 12, 12], [1, 24, 6], [2, 36, 9/4], [3, 48, 3/4], [4, 60, 15/64]], 1359/64, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([12, 24, 36, 48, 60, 72], 1/4, 4, [[0, 12, 12], [1, 24, 6], [2, 36, 9/4], [3, 48, 3/4], [4, 60, 15/64]], 1359/64, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0038","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cee3cc83d267c35f1b940cdb87afb1b3a40e54ef87b0ac48a8d2eff5076fb7b6","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133038,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[13,1],[26,1],[39,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[13, 26, 39] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([13, 26, 39], 1/2, 2, [[0, 13, 13], [1, 26, 13], [2, 39, 39/4]], 143/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([13, 26, 39], 1/2, 2, [[0, 13, 13], [1, 26, 13], [2, 39, 39/4]], 143/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0039","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d83a67d272555c74abaf3f3bb76cd30fd7f14971a48a848d8716d067b90b593","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133039,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[14,1],[28,1],[42,1],[56,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[14, 28, 42, 56] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([14, 28, 42, 56], 1/3, 2, [[0, 14, 14], [1, 28, 28/3], [2, 42, 14/3]], 28, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([14, 28, 42, 56], 1/3, 2, [[0, 14, 14], [1, 28, 28/3], [2, 42, 14/3]], 28, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0040","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"76c8e8a6991ab93d7b522d1d7cdb11df5d3ede454b5088a643540772c87a1991","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133040,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[15,1],[30,1],[45,1],[60,1],[75,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[15, 30, 45, 60, 75] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([15, 30, 45, 60, 75], 1/4, 4, [[0, 15, 15], [1, 30, 15/2], [2, 45, 45/16], [3, 60, 15/16], [4, 75, 75/256]], 6795/256, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([15, 30, 45, 60, 75], 1/4, 4, [[0, 15, 15], [1, 30, 15/2], [2, 45, 45/16], [3, 60, 15/16], [4, 75, 75/256]], 6795/256, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0041","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83429fc984f63afbb343ffa113800c1ba4557825903c51dd599003128ce2789a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133041,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[16,1],[32,1],[48,1],[64,1],[80,1],[96,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[16, 32, 48, 64, 80, 96] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([16, 32, 48, 64, 80, 96], 1/2, 4, [[0, 16, 16], [1, 32, 16], [2, 48, 12], [3, 64, 8], [4, 80, 5]], 57, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([16, 32, 48, 64, 80, 96], 1/2, 4, [[0, 16, 16], [1, 32, 16], [2, 48, 12], [3, 64, 8], [4, 80, 5]], 57, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0042","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"259709572e90f2958662023d141fd1fc438717b4667ce9ad8c1722f3174e14a3","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133042,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[17,1],[34,1],[51,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[17, 34, 51] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([17, 34, 51], 1/3, 2, [[0, 17, 17], [1, 34, 34/3], [2, 51, 17/3]], 34, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([17, 34, 51], 1/3, 2, [[0, 17, 17], [1, 34, 34/3], [2, 51, 17/3]], 34, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0043","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb85758f3c0c2c31a2035e6e36b44b13821bd168a6b9a3a75fc5b450103fec36","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133043,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[18,1],[36,1],[54,1],[72,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[18, 36, 54, 72] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([18, 36, 54, 72], 1/4, 2, [[0, 18, 18], [1, 36, 9], [2, 54, 27/8]], 243/8, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([18, 36, 54, 72], 1/4, 2, [[0, 18, 18], [1, 36, 9], [2, 54, 27/8]], 243/8, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0044","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"642187a612b6916f619ba1785745170e9dd398d9240836f55e8afc15dccaf9cf","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133044,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[19,1],[38,1],[57,1],[76,1],[95,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[19, 38, 57, 76, 95] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([19, 38, 57, 76, 95], 1/2, 4, [[0, 19, 19], [1, 38, 19], [2, 57, 57/4], [3, 76, 19/2], [4, 95, 95/16]], 1083/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([19, 38, 57, 76, 95], 1/2, 4, [[0, 19, 19], [1, 38, 19], [2, 57, 57/4], [3, 76, 19/2], [4, 95, 95/16]], 1083/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0045","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7b9775fbe9602933ec6df73c4ab1fb73ffb3f4e504fe022d137fe352ce8bb06","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133045,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[20,1],[40,1],[60,1],[80,1],[100,1],[120,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[20, 40, 60, 80, 100, 120] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([20, 40, 60, 80, 100, 120], 1/3, 4, [[0, 20, 20], [1, 40, 40/3], [2, 60, 20/3], [3, 80, 80/27], [4, 100, 100/81]], 3580/81, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([20, 40, 60, 80, 100, 120], 1/3, 4, [[0, 20, 20], [1, 40, 40/3], [2, 60, 20/3], [3, 80, 80/27], [4, 100, 100/81]], 3580/81, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0046","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51dbb739a928cd392b55c99a4459cee66ccf1ee4952248e3a1f2a8efe5bbf613","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133046,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[21,1],[42,1],[63,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[21, 42, 63] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([21, 42, 63], 1/4, 2, [[0, 21, 21], [1, 42, 21/2], [2, 63, 63/16]], 567/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([21, 42, 63], 1/4, 2, [[0, 21, 21], [1, 42, 21/2], [2, 63, 63/16]], 567/16, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0047","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"944d3bab3a28f833597dda32b014409fc12e2ed6ffdd0063bc4e33066012fd12","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133047,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[22,1],[44,1],[66,1],[88,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[22, 44, 66, 88] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([22, 44, 66, 88], 1/2, 2, [[0, 22, 22], [1, 44, 22], [2, 66, 33/2]], 121/2, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([22, 44, 66, 88], 1/2, 2, [[0, 22, 22], [1, 44, 22], [2, 66, 33/2]], 121/2, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0048","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"929d4aa95dbc3c60694ab73766b35e978405b8a59852411f2de8d597c6f3fa6a","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133048,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[23,1],[46,1],[69,1],[92,1],[115,1]],\"x\":[1,3]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[23, 46, 69, 92, 115] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/3, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([23, 46, 69, 92, 115], 1/3, 4, [[0, 23, 23], [1, 46, 46/3], [2, 69, 23/3], [3, 92, 92/27], [4, 115, 115/81]], 4117/81, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([23, 46, 69, 92, 115], 1/3, 4, [[0, 23, 23], [1, 46, 46/3], [2, 69, 23/3], [3, 92, 92/27], [4, 115, 115/81]], 4117/81, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0049","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6a58bac4424d6f63438c0af1ee835ea44cc6f33059ab2548677e8f84ac407f3","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133049,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":4,\"finite_support_coefficients\":[[24,1],[48,1],[72,1],[96,1],[120,1],[144,1]],\"x\":[1,4]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[24, 48, 72, 96, 120, 144] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/4, cutoff=4. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([24, 48, 72, 96, 120, 144], 1/4, 4, [[0, 24, 24], [1, 48, 12], [2, 72, 9/2], [3, 96, 3/2], [4, 120, 15/32]], 1359/32, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([24, 48, 72, 96, 120, 144], 1/4, 4, [[0, 24, 24], [1, 48, 12], [2, 72, 9/2], [3, 96, 3/2], [4, 120, 15/32]], 1359/32, No, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0050","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8464a2212e5dd0169eaaed24e2c90368b545dfe1426ed621ac96badf07b83563","family_id":"qf-mod44-ogf-finite-partial-sum","seed":202611133050,"semantic_template":"qf-mod44-ogf-finite-partial-sum/{\"cutoff\":2,\"finite_support_coefficients\":[[25,1],[50,1],[75,1]],\"x\":[1,2]}"},"fields":{"title":"Support Coverage Is Separate from Accidental Equality at One Point","stem":"Use finite_support_coefficients=[25, 50, 75] in increasing index order and the complete law a_n=0 after the list. Canonical answers trim trailing zero coefficients, without changing the original shared parameters. This is a complete finite-support law, not an unknown sequence observed only through a prefix. Use x=1/2, cutoff=2. Give (finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate).","answer":{"kind":"expression","value":"([25, 50, 75], 1/2, 2, [[0, 25, 25], [1, 50, 25], [2, 75, 75/4]], 275/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)","explanation":"List [n,a_n,a_n*x^n] through the cutoff and sum exactly. complete_support_included means every nonzero-support coefficient is included; it is not a pointwise-equality test. At x=0 or through cancellation a shorter sum may have the same value without being the whole polynomial identity. The exact result, retaining the requested order, is ([25, 50, 75], 1/2, 2, [[0, 25, 25], [1, 50, 25], [2, 75, 75/4]], 275/4, Yes, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)."},"skills":["Support Coverage Is Separate from Accidental Equality at One Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0051","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a896b791b9aad257ccf1ca8eb978b95fc1cbb91738c30a464fc1f01cff610fc","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133051,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[1,2],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(1/2)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(1/2)^n_for_every_n>=0, 1/2, 1/4, 4/3, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(1/2)^n_for_every_n>=0, 1/2, 1/4, 4/3, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0052","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3aae6f55a779239a259ec0a807ec8b5f3e6cd57fd534dc323c35ae2261033ce","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133052,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[2,3],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(2/3)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(2/3)^n_for_every_n>=0, 1/2, 1/3, 3/2, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(2/3)^n_for_every_n>=0, 1/2, 1/3, 3/2, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0053","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2324268111af3027846430c88941ee3a12bb3be301a02590d64f612f97badf0","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133053,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[3,4],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(3/4)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(3/4)^n_for_every_n>=0, 1/2, 3/8, 8/5, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(3/4)^n_for_every_n>=0, 1/2, 3/8, 8/5, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0054","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cdc268857183d3191d2fa418fb1da065c83170bdce4824d1d647e3e54597241","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133054,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[4,5],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(4/5)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(4/5)^n_for_every_n>=0, 1/2, 2/5, 5/3, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(4/5)^n_for_every_n>=0, 1/2, 2/5, 5/3, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0055","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43391b1f02b0692f785aa40ee322da44f43e420f857bebcbf6697bd54a26448a","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133055,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[5,6],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(5/6)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(5/6)^n_for_every_n>=0, 1/2, 5/12, 12/7, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(5/6)^n_for_every_n>=0, 1/2, 5/12, 12/7, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0056","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"424c07477b48b981dd486152cd27780c4bd3e136c532025f07fbe0846f1c4890","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133056,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[6,7],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(6/7)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(6/7)^n_for_every_n>=0, 1/2, 3/7, 7/4, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(6/7)^n_for_every_n>=0, 1/2, 3/7, 7/4, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0057","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0edc65c98cb06f01b773b9aa37935c7558563ba15a53316c923c7baf91d27e51","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133057,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[7,8],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(7/8)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(7/8)^n_for_every_n>=0, 1/2, 7/16, 16/9, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(7/8)^n_for_every_n>=0, 1/2, 7/16, 16/9, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0058","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a713ee687cf50b4b62f4f819dcc30855e8d27d8b98348f0dad8cc909dab0967a","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133058,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[8,9],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(8/9)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(8/9)^n_for_every_n>=0, 1/2, 4/9, 9/5, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(8/9)^n_for_every_n>=0, 1/2, 4/9, 9/5, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0059","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e85fcf29ab88901e6a696faf9f669d123657514a168b17ce97619201ae4a227","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133059,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[9,10],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(9/10)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(9/10)^n_for_every_n>=0, 1/2, 9/20, 20/11, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(9/10)^n_for_every_n>=0, 1/2, 9/20, 20/11, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0060","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a812f983c4fe0e13d8b422eeb3ec1e2503b55c02d41f73823b610560b269cb59","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133060,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[10,11],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(10/11)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(10/11)^n_for_every_n>=0, 1/2, 5/11, 11/6, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(10/11)^n_for_every_n>=0, 1/2, 5/11, 11/6, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0061","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7dbaa41294f4f96326854071f112a9208a6c786d86484b4dc5398cab7f51e73","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133061,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[11,12],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(11/12)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(11/12)^n_for_every_n>=0, 1/2, 11/24, 24/13, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(11/12)^n_for_every_n>=0, 1/2, 11/24, 24/13, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0062","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c3ec332639f273b701a327eca37a78f0f3d3129d068c62be0240e10ea6fab44","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133062,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[12,13],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(12/13)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(12/13)^n_for_every_n>=0, 1/2, 6/13, 13/7, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(12/13)^n_for_every_n>=0, 1/2, 6/13, 13/7, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0063","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d0bc06b2468918b1ad4f45e8913238c58f78d0230a260944b405e0f5fefdcac","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133063,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[13,14],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(13/14)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(13/14)^n_for_every_n>=0, 1/2, 13/28, 28/15, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(13/14)^n_for_every_n>=0, 1/2, 13/28, 28/15, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0064","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"262248a8a4aa9cb8de2a417a6c9b05d8cfc0f5283e9fab8499bcaaf74284c343","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133064,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[14,15],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(14/15)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(14/15)^n_for_every_n>=0, 1/2, 7/15, 15/8, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(14/15)^n_for_every_n>=0, 1/2, 7/15, 15/8, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0065","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8131b6275cfd5cc4aa824fc91d4ec7fad36be9ef51c442672063ba6cad80c98","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133065,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[15,16],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(15/16)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(15/16)^n_for_every_n>=0, 1/2, 15/32, 32/17, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(15/16)^n_for_every_n>=0, 1/2, 15/32, 32/17, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0066","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f102204a4264a0f8f459949b7ffc800f5ef46360bc339806c7e7e111aac1c8a","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133066,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[16,17],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(16/17)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(16/17)^n_for_every_n>=0, 1/2, 8/17, 17/9, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(16/17)^n_for_every_n>=0, 1/2, 8/17, 17/9, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0067","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa9ad2706f18b899f848fdd7cd64efa95bb02338931b75d14a67a16d84efec72","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133067,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[17,18],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(17/18)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(17/18)^n_for_every_n>=0, 1/2, 17/36, 36/19, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(17/18)^n_for_every_n>=0, 1/2, 17/36, 36/19, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0068","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"351e9b69db43338204a29449bbd34754f0f48e262a2fdc2c4c4794faac6ff55f","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133068,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[18,19],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(18/19)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(18/19)^n_for_every_n>=0, 1/2, 9/19, 19/10, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(18/19)^n_for_every_n>=0, 1/2, 9/19, 19/10, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0069","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa42ec699f82d339ca5025d816feec63c350d8bcee86b26107e46ba376a50aed","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133069,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[19,20],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(19/20)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(19/20)^n_for_every_n>=0, 1/2, 19/40, 40/21, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(19/20)^n_for_every_n>=0, 1/2, 19/40, 40/21, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0070","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"269fc5548f755ec869f8528ee6e66bf65a498d48a2a486fe220c2d7401d703af","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133070,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[20,21],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(20/21)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(20/21)^n_for_every_n>=0, 1/2, 10/21, 21/11, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(20/21)^n_for_every_n>=0, 1/2, 10/21, 21/11, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0071","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0721e7f19f20b98a5382d911c77be2fe71d65a5aca81bcf65fb52edab7d3f268","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133071,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[21,22],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(21/22)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(21/22)^n_for_every_n>=0, 1/2, 21/44, 44/23, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(21/22)^n_for_every_n>=0, 1/2, 21/44, 44/23, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0072","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86dbaed2c7774250e08abb01d7870e802bbe0131fa89950ae9e2ef5931ad3a31","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133072,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[22,23],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(22/23)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(22/23)^n_for_every_n>=0, 1/2, 11/23, 23/12, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(22/23)^n_for_every_n>=0, 1/2, 11/23, 23/12, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0073","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13d3d7772c42f4957b6ab5c525736b8c169f536749b437a39a4cbf4f3ae43d7d","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133073,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[23,24],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(23/24)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(23/24)^n_for_every_n>=0, 1/2, 23/48, 48/25, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(23/24)^n_for_every_n>=0, 1/2, 23/48, 48/25, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0074","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"362fc619873333d5353cef9896a876b63d3893eb6e5e46969f71d8c12ac64fd3","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133074,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[24,25],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(24/25)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(24/25)^n_for_every_n>=0, 1/2, 12/25, 25/13, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(24/25)^n_for_every_n>=0, 1/2, 12/25, 25/13, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0075","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e66faef0f00907f19ab9247faeb66c732134047703c507b58977e2057f1301d5","family_id":"qf-mod44-ogf-geometric-closed-form","seed":202611133075,"semantic_template":"qf-mod44-ogf-geometric-closed-form/{\"ratio\":[25,26],\"x\":[1,2]}"},"fields":{"title":"The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One","stem":"Use a_n=(25/26)^n for every n≥0 and evaluation x=1/2. Give (coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate).","answer":{"kind":"expression","value":"(a_n=(25/26)^n_for_every_n>=0, 1/2, 25/52, 52/27, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)","explanation":"Require |r*x|<1 before asserting A(x)=1/(1−r*x). A finite denominator alone does not justify analytic equality outside convergence; formal power-series algebra has a separate scope. The exact result, retaining the requested order, is (a_n=(25/26)^n_for_every_n>=0, 1/2, 25/52, 52/27, Yes, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)."},"skills":["The Geometric Analytic Closed Form Requires Absolute r Times x Less Than One"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0076","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7f10c0a0e1c2fcb4f6ddb48da5dffebbddd956eb78ceacde1ecc3b3f9d27d7c","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133076,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[1,1],\"a1\":[2,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=1, a_1=2, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(1, 2, [1, 2, 3, 5, 8, 13, 21, 34, 55, 89], 1/4, 5/4, 11/16, 20/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (1, 2, [1, 2, 3, 5, 8, 13, 21, 34, 55, 89], 1/4, 5/4, 11/16, 20/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0077","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be62f786e616b71ff4f5292c0735763389ad398ecc3f9dea3968818f78a31b88","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133077,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[2,1],\"a1\":[3,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=2, a_1=3, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(2, 3, [2, 3, 5, 8, 13, 21, 34, 55, 89, 144], 1/5, 11/5, 19/25, 55/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (2, 3, [2, 3, 5, 8, 13, 21, 34, 55, 89, 144], 1/5, 11/5, 19/25, 55/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0078","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8d482d7ba1cc6290708ee20b083c3eb5a37e5302fedebc61944a037b781a86f","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133078,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[3,1],\"a1\":[4,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=3, a_1=4, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(3, 4, [3, 4, 7, 11, 18, 29, 47, 76, 123, 199], 1/6, 19/6, 29/36, 114/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (3, 4, [3, 4, 7, 11, 18, 29, 47, 76, 123, 199], 1/6, 19/6, 29/36, 114/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0079","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee43a9c97630c9389d332b68212a072ee44418dcf707a74de72abace772dc576","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133079,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[4,1],\"a1\":[5,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=4, a_1=5, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(4, 5, [4, 5, 9, 14, 23, 37, 60, 97, 157, 254], 1/4, 17/4, 11/16, 68/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (4, 5, [4, 5, 9, 14, 23, 37, 60, 97, 157, 254], 1/4, 17/4, 11/16, 68/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0080","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16e5ac9ed5e30a745f8116cb5b78fe3b05d7f7f651512236684250629002732b","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133080,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[5,1],\"a1\":[6,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=5, a_1=6, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(5, 6, [5, 6, 11, 17, 28, 45, 73, 118, 191, 309], 1/5, 26/5, 19/25, 130/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (5, 6, [5, 6, 11, 17, 28, 45, 73, 118, 191, 309], 1/5, 26/5, 19/25, 130/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0081","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ddf89b8f75799dda45e138ef4203406ca8c1f8e012e1decaef26797522170bd","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133081,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[6,1],\"a1\":[7,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=6, a_1=7, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(6, 7, [6, 7, 13, 20, 33, 53, 86, 139, 225, 364], 1/6, 37/6, 29/36, 222/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (6, 7, [6, 7, 13, 20, 33, 53, 86, 139, 225, 364], 1/6, 37/6, 29/36, 222/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0082","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48f5ec36f4dfec485f7aaf9627a15607af2cd6f8d6dfa9e53641c40f060dc81e","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133082,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[7,1],\"a1\":[8,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=7, a_1=8, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(7, 8, [7, 8, 15, 23, 38, 61, 99, 160, 259, 419], 1/4, 29/4, 11/16, 116/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (7, 8, [7, 8, 15, 23, 38, 61, 99, 160, 259, 419], 1/4, 29/4, 11/16, 116/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0083","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62382902c1cd2facea0176d7adbc149d83a41f8c7f51343ce7e4680bb0ba58e1","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133083,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[8,1],\"a1\":[9,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=8, a_1=9, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(8, 9, [8, 9, 17, 26, 43, 69, 112, 181, 293, 474], 1/5, 41/5, 19/25, 205/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (8, 9, [8, 9, 17, 26, 43, 69, 112, 181, 293, 474], 1/5, 41/5, 19/25, 205/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0084","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1754bbb9411c02ae8a7d124b72f3d4d5077d036ab29fe53322165a74775276d","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133084,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[9,1],\"a1\":[10,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=9, a_1=10, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(9, 10, [9, 10, 19, 29, 48, 77, 125, 202, 327, 529], 1/6, 55/6, 29/36, 330/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (9, 10, [9, 10, 19, 29, 48, 77, 125, 202, 327, 529], 1/6, 55/6, 29/36, 330/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0085","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2dac9d400fae66c889df2f9230680886ee81e40fb525a2870707088a172a3bf8","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133085,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[10,1],\"a1\":[11,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=10, a_1=11, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(10, 11, [10, 11, 21, 32, 53, 85, 138, 223, 361, 584], 1/4, 41/4, 11/16, 164/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (10, 11, [10, 11, 21, 32, 53, 85, 138, 223, 361, 584], 1/4, 41/4, 11/16, 164/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0086","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2df1234d17328ed134bdb08fd835a77fa0e63ba2ed2c78ce5afb18896a4a467","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133086,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[11,1],\"a1\":[12,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=11, a_1=12, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(11, 12, [11, 12, 23, 35, 58, 93, 151, 244, 395, 639], 1/5, 56/5, 19/25, 280/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (11, 12, [11, 12, 23, 35, 58, 93, 151, 244, 395, 639], 1/5, 56/5, 19/25, 280/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0087","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"251e37b0278e51aa26453558471b3ba11d5475e97655830468691e0af7beafa6","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133087,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[12,1],\"a1\":[13,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=12, a_1=13, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(12, 13, [12, 13, 25, 38, 63, 101, 164, 265, 429, 694], 1/6, 73/6, 29/36, 438/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (12, 13, [12, 13, 25, 38, 63, 101, 164, 265, 429, 694], 1/6, 73/6, 29/36, 438/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0088","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5053893c4a6a12d21f38cd7cbf0ecc069c3f21225ae24dbf46be410d4c998531","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133088,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[13,1],\"a1\":[14,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=13, a_1=14, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(13, 14, [13, 14, 27, 41, 68, 109, 177, 286, 463, 749], 1/4, 53/4, 11/16, 212/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (13, 14, [13, 14, 27, 41, 68, 109, 177, 286, 463, 749], 1/4, 53/4, 11/16, 212/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0089","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64e6d3cfa94ed0ae78886f2b0cbb598759341f4825c8009bf2d8b00c59bba164","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133089,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[14,1],\"a1\":[15,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=14, a_1=15, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(14, 15, [14, 15, 29, 44, 73, 117, 190, 307, 497, 804], 1/5, 71/5, 19/25, 355/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (14, 15, [14, 15, 29, 44, 73, 117, 190, 307, 497, 804], 1/5, 71/5, 19/25, 355/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0090","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65c1b5aa56c50159816b6feb0d7394517d87f34765f3de8d5dd1e0165c72985a","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133090,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[15,1],\"a1\":[16,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=15, a_1=16, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(15, 16, [15, 16, 31, 47, 78, 125, 203, 328, 531, 859], 1/6, 91/6, 29/36, 546/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (15, 16, [15, 16, 31, 47, 78, 125, 203, 328, 531, 859], 1/6, 91/6, 29/36, 546/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0091","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbcddb6d254dc17af6fb15b77b231fed093228994f0be9ad76a81310f76caded","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133091,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[16,1],\"a1\":[17,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=16, a_1=17, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(16, 17, [16, 17, 33, 50, 83, 133, 216, 349, 565, 914], 1/4, 65/4, 11/16, 260/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (16, 17, [16, 17, 33, 50, 83, 133, 216, 349, 565, 914], 1/4, 65/4, 11/16, 260/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0092","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b31c7a2aa560d9465081a794678c03ba21468dc62d3bbd4f0c90560a35e4a22","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133092,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[17,1],\"a1\":[18,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=17, a_1=18, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(17, 18, [17, 18, 35, 53, 88, 141, 229, 370, 599, 969], 1/5, 86/5, 19/25, 430/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (17, 18, [17, 18, 35, 53, 88, 141, 229, 370, 599, 969], 1/5, 86/5, 19/25, 430/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0093","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f10bf51c9790f2f88ff1f776a744d133e3fff2d500c43dd359a8589e7e2e6d3","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133093,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[18,1],\"a1\":[19,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=18, a_1=19, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(18, 19, [18, 19, 37, 56, 93, 149, 242, 391, 633, 1024], 1/6, 109/6, 29/36, 654/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (18, 19, [18, 19, 37, 56, 93, 149, 242, 391, 633, 1024], 1/6, 109/6, 29/36, 654/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0094","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2441acc6e305785fbc2751a913d33308a6baf2a6230614dcec16c9824f0caf62","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133094,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[19,1],\"a1\":[20,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=19, a_1=20, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(19, 20, [19, 20, 39, 59, 98, 157, 255, 412, 667, 1079], 1/4, 77/4, 11/16, 28, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (19, 20, [19, 20, 39, 59, 98, 157, 255, 412, 667, 1079], 1/4, 77/4, 11/16, 28, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0095","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24a5dc924e2eef96329e1e23526a2cbd5ae4c889800b6489fe1a09c32fb414a8","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133095,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[20,1],\"a1\":[21,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=20, a_1=21, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(20, 21, [20, 21, 41, 62, 103, 165, 268, 433, 701, 1134], 1/5, 101/5, 19/25, 505/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (20, 21, [20, 21, 41, 62, 103, 165, 268, 433, 701, 1134], 1/5, 101/5, 19/25, 505/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0096","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a11c244b54f19581a23896af3bd73db5d0808ba65ff87423d6aa5851a9e31cff","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133096,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[21,1],\"a1\":[22,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=21, a_1=22, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(21, 22, [21, 22, 43, 65, 108, 173, 281, 454, 735, 1189], 1/6, 127/6, 29/36, 762/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (21, 22, [21, 22, 43, 65, 108, 173, 281, 454, 735, 1189], 1/6, 127/6, 29/36, 762/29, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0097","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd20733412d47921cb773f75c60edde3a26a760859c48aec106c806d1cad61b8","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133097,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[22,1],\"a1\":[23,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=22, a_1=23, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(22, 23, [22, 23, 45, 68, 113, 181, 294, 475, 769, 1244], 1/4, 89/4, 11/16, 356/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (22, 23, [22, 23, 45, 68, 113, 181, 294, 475, 769, 1244], 1/4, 89/4, 11/16, 356/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0098","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b8d6b54b7d47b23730f58bd1ddd71c41c54dd703271b7e2a214b53f480b3056","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133098,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[23,1],\"a1\":[24,1],\"x\":[1,5]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=23, a_1=24, and x=1/5 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(23, 24, [23, 24, 47, 71, 118, 189, 307, 496, 803, 1299], 1/5, 116/5, 19/25, 580/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (23, 24, [23, 24, 47, 71, 118, 189, 307, 496, 803, 1299], 1/5, 116/5, 19/25, 580/19, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0099","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec2b9a5a6e73bd2d554ff24b921754d0f18cc9351941611ecab2fecff3765766","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133099,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[24,1],\"a1\":[25,1],\"x\":[1,6]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=24, a_1=25, and x=1/6 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(24, 25, [24, 25, 49, 74, 123, 197, 320, 517, 837, 1354], 1/6, 145/6, 29/36, 30, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (24, 25, [24, 25, 49, 74, 123, 197, 320, 517, 837, 1354], 1/6, 145/6, 29/36, 30, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r03-analytic-ordinary-generating-function-0100","batch_id":"batch-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"608538bac72950d35609bb5f958df28d202a0b7b131caa18f082b6f6ffde0497","family_id":"qf-mod44-ogf-recurrence-boundary-terms","seed":202611133100,"semantic_template":"qf-mod44-ogf-recurrence-boundary-terms/{\"a0\":[25,1],\"a1\":[26,1],\"x\":[1,4]}"},"fields":{"title":"The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF","stem":"Use the whole recurrence a_n=a_(n−1)+a_(n−2) for every n≥2, initial a_0=25, a_1=26, and x=1/4 with |x|<1/2. Give (a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate).","answer":{"kind":"expression","value":"(25, 26, [25, 26, 51, 77, 128, 205, 333, 538, 871, 1409], 1/4, 101/4, 11/16, 404/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)","explanation":"The ten displayed coefficients are a preview of this complete law, not its definition. Retain numerator a0+(a1−a0)*x and denominator1−x−x^2. The declared domain is sufficient for analytic convergence, not an asserted exact radius for every initial pair. No unproved termwise differentiation or radius-outside substitution is inferred. The exact result, retaining the requested order, is (25, 26, [25, 26, 51, 77, 128, 205, 333, 538, 871, 1409], 1/4, 101/4, 11/16, 404/11, A(x)=(a0+(a1-a0)x)/(1-x-x^2)_with_initial_boundary_terms_preserved)."},"skills":["The Full Recurrence Keeps Both Initial Boundary Terms in Its OGF"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}