遞迴關係基礎 · 脈衝 R05
本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。
本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。
batch-20261014-advance-20261110-r05-recurrence-relation-foundations · 2026-10-14 · 100 題
欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。
欄序與來源綁定資料
1. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[1,2],"constant":-2,"initial_values":[0,1]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [1, 2], -2, [0, 1], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [1, 2], -2, [0, 1], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
2. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[2,4,6],"constant":-1,"initial_values":[1,2,3]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [2, 4, 6], -1, [1, 2, 3], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [2, 4, 6], -1, [1, 2, 3], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
3. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[1,2,3,4],"constant":0,"initial_values":[2,3,4,5]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [1, 2, 3, 4], 0, [2, 3, 4, 5], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [1, 2, 3, 4], 0, [2, 3, 4, 5], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
4. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[2,4],"constant":1,"initial_values":[3,4]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [2, 4], 1, [3, 4], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [2, 4], 1, [3, 4], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
5. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":2,"initial_values":[4,5,6]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [1, 2, 3], 2, [4, 5, 6], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [1, 2, 3], 2, [4, 5, 6], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
6. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[2,4,6,8],"constant":-2,"initial_values":[5,6,7,8]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [2, 4, 6, 8], -2, [5, 6, 7, 8], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [2, 4, 6, 8], -2, [5, 6, 7, 8], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
7. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[1,2],"constant":-1,"initial_values":[6,7]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [1, 2], -1, [6, 7], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [1, 2], -1, [6, 7], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
8. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[2,4,6],"constant":0,"initial_values":[7,8,9]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [2, 4, 6], 0, [7, 8, 9], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [2, 4, 6], 0, [7, 8, 9], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
9. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[1,2,3,4],"constant":1,"initial_values":[8,9,10,11]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [1, 2, 3, 4], 1, [8, 9, 10, 11], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [1, 2, 3, 4], 1, [8, 9, 10, 11], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
10. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[2,4],"constant":2,"initial_values":[9,10]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [2, 4], 2, [9, 10], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [2, 4], 2, [9, 10], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
11. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":-2,"initial_values":[10,11,12]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [1, 2, 3], -2, [10, 11, 12], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [1, 2, 3], -2, [10, 11, 12], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
12. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[2,4,6,8],"constant":-1,"initial_values":[11,12,13,14]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [2, 4, 6, 8], -1, [11, 12, 13, 14], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [2, 4, 6, 8], -1, [11, 12, 13, 14], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
13. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[1,2],"constant":0,"initial_values":[12,13]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [1, 2], 0, [12, 13], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [1, 2], 0, [12, 13], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
14. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[2,4,6],"constant":1,"initial_values":[13,14,15]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [2, 4, 6], 1, [13, 14, 15], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [2, 4, 6], 1, [13, 14, 15], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
15. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[1,2,3,4],"constant":2,"initial_values":[14,15,16,17]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [1, 2, 3, 4], 2, [14, 15, 16, 17], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [1, 2, 3, 4], 2, [14, 15, 16, 17], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
16. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[2,4],"constant":-2,"initial_values":[15,16]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [2, 4], -2, [15, 16], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [2, 4], -2, [15, 16], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
17. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":-1,"initial_values":[16,17,18]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [1, 2, 3], -1, [16, 17, 18], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [1, 2, 3], -1, [16, 17, 18], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
18. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[2,4,6,8],"constant":0,"initial_values":[17,18,19,20]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [2, 4, 6, 8], 0, [17, 18, 19, 20], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [2, 4, 6, 8], 0, [17, 18, 19, 20], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
19. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[1,2],"constant":1,"initial_values":[18,19]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [1, 2], 1, [18, 19], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [1, 2], 1, [18, 19], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
20. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[2,4,6],"constant":2,"initial_values":[19,20,21]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [2, 4, 6], 2, [19, 20, 21], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [2, 4, 6], 2, [19, 20, 21], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
21. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[1,2,3,4],"constant":-2,"initial_values":[20,21,22,23]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [1, 2, 3, 4], -2, [20, 21, 22, 23], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [1, 2, 3, 4], -2, [20, 21, 22, 23], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
22. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[2,4],"constant":-1,"initial_values":[21,22]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [2, 4], -1, [21, 22], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [2, 4], -1, [21, 22], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
23. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":0,"initial_values":[22,23,24]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(3, [1, 2, 3], 0, [22, 23, 24], 是, 是, rule_applies_for_n_ge_3)
Wave6第四層核對=(3, [1, 2, 3], 0, [22, 23, 24], 是, 是, rule_applies_for_n_ge_3)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
24. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":4,"coefficients":[2,4,6,8],"constant":1,"initial_values":[23,24,25,26]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(4, [2, 4, 6, 8], 1, [23, 24, 25, 26], 是, 是, rule_applies_for_n_ge_4)
Wave6第四層核對=(4, [2, 4, 6, 8], 1, [23, 24, 25, 26], 是, 是, rule_applies_for_n_ge_4)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
25. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":2,"coefficients":[1,2],"constant":2,"initial_values":[24,25]}。請依 [order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope] 欄序回答。
查看答案與解釋
(2, [1, 2], 2, [24, 25], 是, 是, rule_applies_for_n_ge_2)
Wave6第四層核對=(2, [1, 2], 2, [24, 25], 是, 是, rule_applies_for_n_ge_2)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-specification-qualification · candidate
26. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[1,2],"constant":-2,"initial_values":[0,1],"target_n":5}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [0, 1], 5, [0, 1, 0, -1, -4, -11], -11, exact_forward_application_only)
Wave6第四層核對=(2, [0, 1], 5, [0, 1, 0, -1, -4, -11], -11, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
27. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[2,4,6],"constant":-1,"initial_values":[1,2,3],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [1, 2, 3], 7, [1, 2, 3, 27, 177, 1175, 7811, 51919], 51919, exact_forward_application_only)
Wave6第四層核對=(3, [1, 2, 3], 7, [1, 2, 3, 27, 177, 1175, 7811, 51919], 51919, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
28. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[1,2,3,4],"constant":0,"initial_values":[2,3,4,5],"target_n":9}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [2, 3, 4, 5], 9, [2, 3, 4, 5, 40, 186, 878, 4155, 19666, 93071], 93071, exact_forward_application_only)
Wave6第四層核對=(4, [2, 3, 4, 5], 9, [2, 3, 4, 5, 40, 186, 878, 4155, 19666, 93071], 93071, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
29. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[2,4],"constant":1,"initial_values":[3,4],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [3, 4], 8, [3, 4, 23, 101, 451, 2007, 8931, 39739, 176819], 176819, exact_forward_application_only)
Wave6第四層核對=(2, [3, 4], 8, [3, 4, 23, 101, 451, 2007, 8931, 39739, 176819], 176819, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
30. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[1,2,3],"constant":2,"initial_values":[4,5,6],"target_n":6}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [4, 5, 6], 6, [4, 5, 6, 34, 121, 439, 1595], 1595, exact_forward_application_only)
Wave6第四層核對=(3, [4, 5, 6], 6, [4, 5, 6, 34, 121, 439, 1595], 1595, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
31. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[2,4,6,8],"constant":-2,"initial_values":[5,6,7,8],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [5, 6, 7, 8], 8, [5, 6, 7, 8, 138, 1190, 10392, 90842, 794122], 794122, exact_forward_application_only)
Wave6第四層核對=(4, [5, 6, 7, 8], 8, [5, 6, 7, 8, 138, 1190, 10392, 90842, 794122], 794122, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
32. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[1,2],"constant":-1,"initial_values":[6,7],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [6, 7], 7, [6, 7, 19, 44, 106, 255, 615, 1484], 1484, exact_forward_application_only)
Wave6第四層核對=(2, [6, 7], 7, [6, 7, 19, 44, 106, 255, 615, 1484], 1484, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
33. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[2,4,6],"constant":0,"initial_values":[7,8,9],"target_n":9}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [7, 8, 9], 9, [7, 8, 9, 100, 652, 4330, 28788, 191352, 1271924, 8454528], 8454528, exact_forward_application_only)
Wave6第四層核對=(3, [7, 8, 9], 9, [7, 8, 9, 100, 652, 4330, 28788, 191352, 1271924, 8454528], 8454528, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
34. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[1,2,3,4],"constant":1,"initial_values":[8,9,10,11],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [8, 9, 10, 11], 7, [8, 9, 10, 11, 101, 467, 2204, 10431], 10431, exact_forward_application_only)
Wave6第四層核對=(4, [8, 9, 10, 11], 7, [8, 9, 10, 11, 101, 467, 2204, 10431], 10431, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
35. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[2,4],"constant":2,"initial_values":[9,10],"target_n":6}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [9, 10], 6, [9, 10, 60, 262, 1170, 5206, 23166], 23166, exact_forward_application_only)
Wave6第四層核對=(2, [9, 10], 6, [9, 10, 60, 262, 1170, 5206, 23166], 23166, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
36. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[1,2,3],"constant":-2,"initial_values":[10,11,12],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [10, 11, 12], 8, [10, 11, 12, 66, 231, 835, 3031, 10992, 39871], 39871, exact_forward_application_only)
Wave6第四層核對=(3, [10, 11, 12], 8, [10, 11, 12, 66, 231, 835, 3031, 10992, 39871], 39871, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
37. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[2,4,6,8],"constant":-1,"initial_values":[11,12,13,14],"target_n":10}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [11, 12, 13, 14], 10, [11, 12, 13, 14, 259, 2231, 19483, 170313, 1488843, 13015015, 113773395], 113773395, exact_forward_application_only)
Wave6第四層核對=(4, [11, 12, 13, 14], 10, [11, 12, 13, 14, 259, 2231, 19483, 170313, 1488843, 13015015, 113773395], 113773395, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
38. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[1,2],"constant":0,"initial_values":[12,13],"target_n":5}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [12, 13], 5, [12, 13, 38, 89, 216, 521], 521, exact_forward_application_only)
Wave6第四層核對=(2, [12, 13], 5, [12, 13, 38, 89, 216, 521], 521, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
39. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[2,4,6],"constant":1,"initial_values":[13,14,15],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [13, 14, 15], 7, [13, 14, 15, 173, 1127, 7485, 49765, 330785], 330785, exact_forward_application_only)
Wave6第四層核對=(3, [13, 14, 15], 7, [13, 14, 15, 173, 1127, 7485, 49765, 330785], 330785, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
40. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[1,2,3,4],"constant":2,"initial_values":[14,15,16,17],"target_n":9}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [14, 15, 16, 17], 9, [14, 15, 16, 17, 162, 748, 3530, 16707, 79078, 374243], 374243, exact_forward_application_only)
Wave6第四層核對=(4, [14, 15, 16, 17], 9, [14, 15, 16, 17, 162, 748, 3530, 16707, 79078, 374243], 374243, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
41. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[2,4],"constant":-2,"initial_values":[15,16],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [15, 16], 8, [15, 16, 92, 398, 1774, 7890, 35106, 156202, 695018], 695018, exact_forward_application_only)
Wave6第四層核對=(2, [15, 16], 8, [15, 16, 92, 398, 1774, 7890, 35106, 156202, 695018], 695018, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
42. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[1,2,3],"constant":-1,"initial_values":[16,17,18],"target_n":6}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [16, 17, 18], 6, [16, 17, 18, 103, 361, 1306, 4742], 4742, exact_forward_application_only)
Wave6第四層核對=(3, [16, 17, 18], 6, [16, 17, 18, 103, 361, 1306, 4742], 4742, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
43. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[2,4,6,8],"constant":0,"initial_values":[17,18,19,20],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [17, 18, 19, 20], 8, [17, 18, 19, 20, 380, 3272, 28574, 249784, 2183564], 2183564, exact_forward_application_only)
Wave6第四層核對=(4, [17, 18, 19, 20], 8, [17, 18, 19, 20, 380, 3272, 28574, 249784, 2183564], 2183564, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
44. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[1,2],"constant":1,"initial_values":[18,19],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [18, 19], 7, [18, 19, 57, 134, 326, 787, 1901, 4590], 4590, exact_forward_application_only)
Wave6第四層核對=(2, [18, 19], 7, [18, 19, 57, 134, 326, 787, 1901, 4590], 4590, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
45. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[2,4,6],"constant":2,"initial_values":[19,20,21],"target_n":9}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [19, 20, 21], 9, [19, 20, 21, 246, 1602, 10640, 70742, 470218, 3125558, 20775706], 20775706, exact_forward_application_only)
Wave6第四層核對=(3, [19, 20, 21], 9, [19, 20, 21, 246, 1602, 10640, 70742, 470218, 3125558, 20775706], 20775706, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
46. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[1,2,3,4],"constant":-2,"initial_values":[20,21,22,23],"target_n":7}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [20, 21, 22, 23], 7, [20, 21, 22, 23, 218, 1004, 4736, 22413], 22413, exact_forward_application_only)
Wave6第四層核對=(4, [20, 21, 22, 23], 7, [20, 21, 22, 23, 218, 1004, 4736, 22413], 22413, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
47. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[2,4],"constant":-1,"initial_values":[21,22],"target_n":6}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [21, 22], 6, [21, 22, 129, 559, 2493, 11089, 49341], 49341, exact_forward_application_only)
Wave6第四層核對=(2, [21, 22], 6, [21, 22, 129, 559, 2493, 11089, 49341], 49341, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
48. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":3,"coefficients":[1,2,3],"constant":0,"initial_values":[22,23,24],"target_n":8}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(3, [22, 23, 24], 8, [22, 23, 24, 140, 491, 1777, 6453, 23404, 84895], 84895, exact_forward_application_only)
Wave6第四層核對=(3, [22, 23, 24], 8, [22, 23, 24, 140, 491, 1777, 6453, 23404, 84895], 84895, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
49. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":4,"coefficients":[2,4,6,8],"constant":1,"initial_values":[23,24,25,26],"target_n":10}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(4, [23, 24, 25, 26], 10, [23, 24, 25, 26, 501, 4313, 37665, 329255, 2878285, 25161097, 219950837], 219950837, exact_forward_application_only)
Wave6第四層核對=(4, [23, 24, 25, 26], 10, [23, 24, 25, 26, 501, 4313, 37665, 329255, 2878285, 25161097, 219950837], 219950837, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
50. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-forward-generation；資料={"order":2,"coefficients":[1,2],"constant":2,"initial_values":[24,25],"target_n":5}。請依 [order, initial_values, target_n, prefix, target_value, certificate] 欄序回答。
查看答案與解釋
(2, [24, 25], 5, [24, 25, 76, 179, 436, 1053], 1053, exact_forward_application_only)
Wave6第四層核對=(2, [24, 25], 5, [24, 25, 76, 179, 436, 1053], 1053, exact_forward_application_only)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-forward-generation · candidate
51. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[1,2],"constant":-2,"initial_values":[0,1],"target_n":5}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 5, -11, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 5, -11, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
52. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[2,4,6],"constant":-1,"initial_values":[1,2,3],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 7, 51919, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 7, 51919, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
53. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[1,2,3,4],"constant":0,"initial_values":[2,3,4,5],"target_n":9}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 9, 93071, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 9, 93071, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
54. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[2,4],"constant":1,"initial_values":[3,4],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 8, 176819, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 8, 176819, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
55. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[1,2,3],"constant":2,"initial_values":[4,5,6],"target_n":6}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 6, 1595, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 6, 1595, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
56. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[2,4,6,8],"constant":-2,"initial_values":[5,6,7,8],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 8, 794122, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 8, 794122, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
57. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[1,2],"constant":-1,"initial_values":[6,7],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 7, 1484, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 7, 1484, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
58. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[2,4,6],"constant":0,"initial_values":[7,8,9],"target_n":9}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 9, 8454528, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 9, 8454528, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
59. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[1,2,3,4],"constant":1,"initial_values":[8,9,10,11],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 7, 10431, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 7, 10431, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
60. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[2,4],"constant":2,"initial_values":[9,10],"target_n":6}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 6, 23166, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 6, 23166, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
61. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[1,2,3],"constant":-2,"initial_values":[10,11,12],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 8, 39871, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 8, 39871, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
62. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[2,4,6,8],"constant":-1,"initial_values":[11,12,13,14],"target_n":10}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 10, 113773395, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 10, 113773395, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
63. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[1,2],"constant":0,"initial_values":[12,13],"target_n":5}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 5, 521, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 5, 521, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
64. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[2,4,6],"constant":1,"initial_values":[13,14,15],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 7, 330785, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 7, 330785, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
65. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[1,2,3,4],"constant":2,"initial_values":[14,15,16,17],"target_n":9}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 9, 374243, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 9, 374243, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
66. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[2,4],"constant":-2,"initial_values":[15,16],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 8, 695018, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 8, 695018, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
67. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[1,2,3],"constant":-1,"initial_values":[16,17,18],"target_n":6}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 6, 4742, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 6, 4742, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
68. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[2,4,6,8],"constant":0,"initial_values":[17,18,19,20],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 8, 2183564, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 8, 2183564, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
69. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[1,2],"constant":1,"initial_values":[18,19],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 7, 4590, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 7, 4590, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
70. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[2,4,6],"constant":2,"initial_values":[19,20,21],"target_n":9}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 9, 20775706, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 9, 20775706, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
71. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[1,2,3,4],"constant":-2,"initial_values":[20,21,22,23],"target_n":7}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 7, 22413, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 7, 22413, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
72. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[2,4],"constant":-1,"initial_values":[21,22],"target_n":6}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 6, 49341, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 6, 49341, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
73. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":3,"coefficients":[1,2,3],"constant":0,"initial_values":[22,23,24],"target_n":8}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(3, 3, 8, 84895, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(3, 3, 8, 84895, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
74. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":4,"coefficients":[2,4,6,8],"constant":1,"initial_values":[23,24,25,26],"target_n":10}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(4, 4, 10, 219950837, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(4, 4, 10, 219950837, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
75. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-initial-value-uniqueness；資料={"order":2,"coefficients":[1,2],"constant":2,"initial_values":[24,25],"target_n":5}。請依 [order, initial_count, target_n, target_value, unique_prefix, certificate] 欄序回答。
查看答案與解釋
(2, 2, 5, 1053, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)
Wave6第四層核對=(2, 2, 5, 1053, 是, total_single_valued_rule_plus_all_order_many_initial_values_gives_unique_prefix)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-initial-value-uniqueness · candidate
76. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,2],"constant":0,"first_initial":0,"second_left":1,"second_right":2,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(0, 1, 2, [0, 1, 2, 5, 12, 29, 70], [0, 2, 4, 10, 24, 58, 140], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(0, 1, 2, [0, 1, 2, 5, 12, 29, 70], [0, 2, 4, 10, 24, 58, 140], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
77. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,3],"constant":1,"first_initial":1,"second_left":2,"second_right":3,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(1, 2, 3, [1, 2, 9, 32, 115, 410, 1461, 5204], [1, 3, 12, 43, 154, 549, 1956, 6967], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(1, 2, 3, [1, 2, 9, 32, 115, 410, 1461, 5204], [1, 3, 12, 43, 154, 549, 1956, 6967], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
78. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,4],"constant":2,"first_initial":2,"second_left":3,"second_right":4,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(2, 3, 4, [2, 3, 20, 91, 426, 1979, 9196, 42723, 198482], [2, 4, 24, 110, 514, 2388, 11096, 51550, 239490], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(2, 3, 4, [2, 3, 20, 91, 426, 1979, 9196, 42723, 198482], [2, 4, 24, 110, 514, 2388, 11096, 51550, 239490], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
79. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,5],"constant":3,"first_initial":3,"second_left":4,"second_right":5,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(3, 4, 5, [3, 4, 26, 137, 714, 3710, 19267], [3, 5, 31, 163, 849, 4411, 22907], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(3, 4, 5, [3, 4, 26, 137, 714, 3710, 19267], [3, 5, 31, 163, 849, 4411, 22907], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
80. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,6],"constant":0,"first_initial":4,"second_left":5,"second_right":6,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(4, 5, 6, [4, 5, 38, 238, 1504, 9500, 60008, 379048], [4, 6, 44, 276, 1744, 11016, 69584, 439536], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(4, 5, 6, [4, 5, 38, 238, 1504, 9500, 60008, 379048], [4, 6, 44, 276, 1744, 11016, 69584, 439536], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
81. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,2],"constant":1,"first_initial":5,"second_left":6,"second_right":7,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(5, 6, 7, [5, 6, 28, 75, 235, 696, 2098, 6285, 18865], [5, 7, 30, 82, 255, 757, 2280, 6832, 20505], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(5, 6, 7, [5, 6, 28, 75, 235, 696, 2098, 6285, 18865], [5, 7, 30, 82, 255, 757, 2280, 6832, 20505], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
82. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,3],"constant":2,"first_initial":6,"second_left":7,"second_right":8,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(6, 7, 8, [6, 7, 29, 96, 319, 1055, 3486], [6, 8, 32, 106, 352, 1164, 3846], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(6, 7, 8, [6, 7, 29, 96, 319, 1055, 3486], [6, 8, 32, 106, 352, 1164, 3846], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
83. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,4],"constant":3,"first_initial":7,"second_left":8,"second_right":9,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(7, 8, 9, [7, 8, 49, 215, 961, 4277, 19033, 84689], [7, 9, 53, 233, 1041, 4633, 20617, 91737], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(7, 8, 9, [7, 8, 49, 215, 961, 4277, 19033, 84689], [7, 9, 53, 233, 1041, 4633, 20617, 91737], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
84. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,5],"constant":0,"first_initial":8,"second_left":9,"second_right":10,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(8, 9, 10, [8, 9, 69, 372, 2067, 11451, 63456, 351633, 1948533], [8, 10, 74, 400, 2222, 12310, 68216, 378010, 2094698], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(8, 9, 10, [8, 9, 69, 372, 2067, 11451, 63456, 351633, 1948533], [8, 10, 74, 400, 2222, 12310, 68216, 378010, 2094698], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
85. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,6],"constant":1,"first_initial":9,"second_left":10,"second_right":11,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(9, 10, 11, [9, 10, 70, 431, 2657, 16374, 100902], [9, 11, 76, 468, 2885, 17779, 109560], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(9, 10, 11, [9, 10, 70, 431, 2657, 16374, 100902], [9, 11, 76, 468, 2885, 17779, 109560], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
86. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,2],"constant":2,"first_initial":10,"second_left":11,"second_right":12,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(10, 11, 12, [10, 11, 44, 112, 314, 854, 2338, 6386], [10, 12, 46, 118, 330, 898, 2458, 6714], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(10, 11, 12, [10, 11, 44, 112, 314, 854, 2338, 6386], [10, 12, 46, 118, 330, 898, 2458, 6714], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
87. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,3],"constant":3,"first_initial":11,"second_left":12,"second_right":13,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(11, 12, 13, [11, 12, 72, 255, 984, 3720, 14115, 53508, 202872], [11, 13, 75, 267, 1029, 3891, 14763, 55965, 212187], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(11, 12, 13, [11, 12, 72, 255, 984, 3720, 14115, 53508, 202872], [11, 13, 75, 267, 1029, 3891, 14763, 55965, 212187], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
88. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,4],"constant":0,"first_initial":12,"second_left":13,"second_right":14,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(12, 13, 14, [12, 13, 64, 269, 1140, 4829, 20456], [12, 14, 68, 286, 1212, 5134, 21748], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(12, 13, 14, [12, 13, 64, 269, 1140, 4829, 20456], [12, 14, 68, 286, 1212, 5134, 21748], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
89. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,5],"constant":1,"first_initial":13,"second_left":14,"second_right":15,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(13, 14, 15, [13, 14, 97, 514, 2765, 14854, 79801, 428714], [13, 15, 102, 541, 2910, 15633, 83986, 451197], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(13, 14, 15, [13, 14, 97, 514, 2765, 14854, 79801, 428714], [13, 15, 102, 541, 2910, 15633, 83986, 451197], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
90. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,6],"constant":2,"first_initial":14,"second_left":15,"second_right":16,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(14, 15, 16, [14, 15, 134, 851, 5510, 35615, 230222, 1488179, 9619742], [14, 16, 140, 890, 5762, 37244, 240752, 1556246, 10059734], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(14, 15, 16, [14, 15, 134, 851, 5510, 35615, 230222, 1488179, 9619742], [14, 16, 140, 890, 5762, 37244, 240752, 1556246, 10059734], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
91. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,2],"constant":3,"first_initial":15,"second_left":16,"second_right":17,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(15, 16, 17, [15, 16, 50, 119, 291, 704, 1702], [15, 17, 52, 124, 303, 733, 1772], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(15, 16, 17, [15, 16, 50, 119, 291, 704, 1702], [15, 17, 52, 124, 303, 733, 1772], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
92. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,3],"constant":0,"first_initial":16,"second_left":17,"second_right":18,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(16, 17, 18, [16, 17, 83, 283, 1015, 3611, 12863, 45811], [16, 18, 86, 294, 1054, 3750, 13358, 47574], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(16, 17, 18, [16, 17, 83, 283, 1015, 3611, 12863, 45811], [16, 18, 86, 294, 1054, 3750, 13358, 47574], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
93. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,4],"constant":1,"first_initial":17,"second_left":18,"second_right":19,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(17, 18, 19, [17, 18, 124, 551, 2577, 11962, 55580, 258207, 1199569], [17, 19, 128, 570, 2665, 12371, 57480, 267034, 1240577], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(17, 18, 19, [17, 18, 124, 551, 2577, 11962, 55580, 258207, 1199569], [17, 19, 128, 570, 2665, 12371, 57480, 267034, 1240577], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
94. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,5],"constant":2,"first_initial":18,"second_left":19,"second_right":20,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(18, 19, 20, [18, 19, 115, 596, 3097, 16083, 83514], [18, 20, 120, 622, 3232, 16784, 87154], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(18, 19, 20, [18, 19, 115, 596, 3097, 16083, 83514], [18, 20, 120, 622, 3232, 16784, 87154], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
95. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,6],"constant":3,"first_initial":19,"second_left":20,"second_right":21,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(19, 20, 21, [19, 20, 161, 1009, 6379, 40295, 254531, 1607779], [19, 21, 167, 1047, 6619, 41811, 264107, 1668267], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(19, 20, 21, [19, 20, 161, 1009, 6379, 40295, 254531, 1607779], [19, 21, 167, 1047, 6619, 41811, 264107, 1668267], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
96. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,2],"constant":0,"first_initial":20,"second_left":21,"second_right":22,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(20, 21, 22, [20, 21, 102, 267, 840, 2481, 7482, 22407, 67260], [20, 22, 104, 274, 860, 2542, 7664, 22954, 68900], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(20, 21, 22, [20, 21, 102, 267, 840, 2481, 7482, 22407, 67260], [20, 22, 104, 274, 860, 2542, 7664, 22954, 68900], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
97. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,3],"constant":1,"first_initial":21,"second_left":22,"second_right":23,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(21, 22, 23, [21, 22, 88, 287, 950, 3138, 10365], [21, 23, 91, 297, 983, 3247, 10725], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(21, 22, 23, [21, 22, 88, 287, 950, 3138, 10365], [21, 23, 91, 297, 983, 3247, 10725], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
98. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[2,4],"constant":2,"first_initial":22,"second_left":23,"second_right":24,"target_n":7}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(22, 23, 24, [22, 23, 138, 600, 2678, 11914, 53014, 235886], [22, 24, 142, 618, 2758, 12270, 54598, 242934], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(22, 23, 24, [22, 23, 138, 600, 2678, 11914, 53014, 235886], [22, 24, 142, 618, 2758, 12270, 54598, 242934], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
99. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[3,5],"constant":3,"first_initial":23,"second_left":24,"second_right":25,"target_n":8}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(23, 24, 25, [23, 24, 192, 1035, 5754, 31878, 176655, 978912, 5424528], [23, 25, 197, 1063, 5909, 32737, 181415, 1005289, 5570693], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(23, 24, 25, [23, 24, 192, 1035, 5754, 31878, 176655, 978912, 5424528], [23, 25, 197, 1063, 5909, 32737, 181415, 1005289, 5570693], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
100. 遞迴關係
initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-missing-initial-boundary；資料={"coefficients":[1,6],"constant":0,"first_initial":24,"second_left":25,"second_right":26,"target_n":6}。請依 [first_initial, second_left, second_right, left_prefix, right_prefix, prefixes_differ, one_initial_uniqueness_claim, boundary] 欄序回答。
prefixes_differ比較包含初值的完整prefix；不保證兩種初值在指定target_n的值仍不同。
查看答案與解釋
(24, 25, 26, [24, 25, 174, 1069, 6588, 40597, 250170], [24, 26, 180, 1106, 6816, 42002, 258828], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)
Wave6第四層核對=(24, 25, 26, [24, 25, 174, 1069, 6588, 40597, 250170], [24, 26, 180, 1106, 6816, 42002, 258828], 是, 否, order_two_recurrence_with_only_one_initial_value_is_not_unique)；依完整明示資料獨立重算；state shift-left後append新scalar term；order-r recurrence須有r個必要初值才唯一決定forward prefix；有限生成prefix不證closed form、收斂或漸近性；缺初值或多值rule不得宣稱unique。
本題逐步詳解（AI 二次覆核）
遞迴關係 · qf-mod41-recurrence-missing-initial-boundary · candidate
原始題目 JSONL · 驗收報告