解析普通生成函數基礎 · 脈衝 R03

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-20261015-advance-20261113-r03-analytic-ordinary-generating-function-foundations · 2026-10-15 · 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. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[1,2,3],"index":0}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([1, 2, 3], a_n=0_after_the_list, 0, 1, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([1, 2, 3], a_n=0_after_the_list, 0, 1, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

2. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[2,4,6,8],"index":1}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([2, 4, 6, 8], a_n=0_after_the_list, 1, 4, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([2, 4, 6, 8], a_n=0_after_the_list, 1, 4, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

3. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[3,6,9,12,15],"index":2}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([3, 6, 9, 12, 15], a_n=0_after_the_list, 2, 9, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([3, 6, 9, 12, 15], a_n=0_after_the_list, 2, 9, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

4. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[4,8,12,16,20,24],"index":3}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([4, 8, 12, 16, 20, 24], a_n=0_after_the_list, 3, 16, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([4, 8, 12, 16, 20, 24], a_n=0_after_the_list, 3, 16, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

5. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[5,10,15],"index":4}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([5, 10, 15], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([5, 10, 15], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

6. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[6,12,18,24],"index":5}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([6, 12, 18, 24], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([6, 12, 18, 24], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

7. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[7,14,21,28,35],"index":6}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([7, 14, 21, 28, 35], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([7, 14, 21, 28, 35], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

8. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[8,16,24,32,40,48],"index":7}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([8, 16, 24, 32, 40, 48], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([8, 16, 24, 32, 40, 48], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

9. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[9,18,27],"index":8}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([9, 18, 27], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([9, 18, 27], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

10. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[10,20,30,40],"index":0}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([10, 20, 30, 40], a_n=0_after_the_list, 0, 10, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([10, 20, 30, 40], a_n=0_after_the_list, 0, 10, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

11. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[11,22,33,44,55],"index":1}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([11, 22, 33, 44, 55], a_n=0_after_the_list, 1, 22, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([11, 22, 33, 44, 55], a_n=0_after_the_list, 1, 22, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

12. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[12,24,36,48,60,72],"index":2}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([12, 24, 36, 48, 60, 72], a_n=0_after_the_list, 2, 36, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([12, 24, 36, 48, 60, 72], a_n=0_after_the_list, 2, 36, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

13. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[13,26,39],"index":3}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([13, 26, 39], a_n=0_after_the_list, 3, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([13, 26, 39], a_n=0_after_the_list, 3, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

14. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[14,28,42,56],"index":4}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([14, 28, 42, 56], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([14, 28, 42, 56], a_n=0_after_the_list, 4, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

15. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[15,30,45,60,75],"index":5}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([15, 30, 45, 60, 75], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([15, 30, 45, 60, 75], a_n=0_after_the_list, 5, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

16. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[16,32,48,64,80,96],"index":6}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([16, 32, 48, 64, 80, 96], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([16, 32, 48, 64, 80, 96], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

17. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[17,34,51],"index":7}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([17, 34, 51], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([17, 34, 51], a_n=0_after_the_list, 7, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

18. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[18,36,54,72],"index":8}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([18, 36, 54, 72], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([18, 36, 54, 72], a_n=0_after_the_list, 8, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

19. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[19,38,57,76,95],"index":0}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([19, 38, 57, 76, 95], a_n=0_after_the_list, 0, 19, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([19, 38, 57, 76, 95], a_n=0_after_the_list, 0, 19, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

20. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[20,40,60,80,100,120],"index":1}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([20, 40, 60, 80, 100, 120], a_n=0_after_the_list, 1, 40, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([20, 40, 60, 80, 100, 120], a_n=0_after_the_list, 1, 40, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

21. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[21,42,63],"index":2}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([21, 42, 63], a_n=0_after_the_list, 2, 63, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([21, 42, 63], a_n=0_after_the_list, 2, 63, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

22. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[22,44,66,88],"index":3}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([22, 44, 66, 88], a_n=0_after_the_list, 3, 88, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([22, 44, 66, 88], a_n=0_after_the_list, 3, 88, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

23. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[23,46,69,92,115],"index":4}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([23, 46, 69, 92, 115], a_n=0_after_the_list, 4, 115, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([23, 46, 69, 92, 115], a_n=0_after_the_list, 4, 115, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

24. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[24,48,72,96,120,144],"index":5}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([24, 48, 72, 96, 120, 144], a_n=0_after_the_list, 5, 144, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([24, 48, 72, 96, 120, 144], a_n=0_after_the_list, 5, 144, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

25. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-coefficient-extraction;資料={"finite_support_coefficients":[25,50,75],"index":6}。請依 [finite_support_coefficients, zero_tail_law, index, coefficient, certificate] 欄序回答。

查看答案與解釋

([25, 50, 75], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law)

Wave6第七層核對=([25, 50, 75], a_n=0_after_the_list, 6, 0, coefficient_extraction_uses_the_complete_finite_support_law);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-coefficient-extraction · candidate

26. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[1,2,3],"x":[1,2],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([1, 2, 3], 1/2, 2, [[0, 1, 1], [1, 2, 1], [2, 3, 3/4]], 11/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([1, 2, 3], 1/2, 2, [[0, 1, 1], [1, 2, 1], [2, 3, 3/4]], 11/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

27. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[2,4,6,8],"x":[1,3],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([2, 4, 6, 8], 1/3, 2, [[0, 2, 2], [1, 4, 4/3], [2, 6, 2/3]], 4, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([2, 4, 6, 8], 1/3, 2, [[0, 2, 2], [1, 4, 4/3], [2, 6, 2/3]], 4, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

28. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[3,6,9,12,15],"x":[1,4],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

29. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[4,8,12,16,20,24],"x":[1,2],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

30. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[5,10,15],"x":[1,3],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([5, 10, 15], 1/3, 2, [[0, 5, 5], [1, 10, 10/3], [2, 15, 5/3]], 10, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([5, 10, 15], 1/3, 2, [[0, 5, 5], [1, 10, 10/3], [2, 15, 5/3]], 10, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

31. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[6,12,18,24],"x":[1,4],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([6, 12, 18, 24], 1/4, 2, [[0, 6, 6], [1, 12, 3], [2, 18, 9/8]], 81/8, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([6, 12, 18, 24], 1/4, 2, [[0, 6, 6], [1, 12, 3], [2, 18, 9/8]], 81/8, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

32. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[7,14,21,28,35],"x":[1,2],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

33. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[8,16,24,32,40,48],"x":[1,3],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

34. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[9,18,27],"x":[1,4],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([9, 18, 27], 1/4, 2, [[0, 9, 9], [1, 18, 9/2], [2, 27, 27/16]], 243/16, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([9, 18, 27], 1/4, 2, [[0, 9, 9], [1, 18, 9/2], [2, 27, 27/16]], 243/16, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

35. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[10,20,30,40],"x":[1,2],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([10, 20, 30, 40], 1/2, 2, [[0, 10, 10], [1, 20, 10], [2, 30, 15/2]], 55/2, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([10, 20, 30, 40], 1/2, 2, [[0, 10, 10], [1, 20, 10], [2, 30, 15/2]], 55/2, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

36. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[11,22,33,44,55],"x":[1,3],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

37. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[12,24,36,48,60,72],"x":[1,4],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

38. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[13,26,39],"x":[1,2],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([13, 26, 39], 1/2, 2, [[0, 13, 13], [1, 26, 13], [2, 39, 39/4]], 143/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([13, 26, 39], 1/2, 2, [[0, 13, 13], [1, 26, 13], [2, 39, 39/4]], 143/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

39. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[14,28,42,56],"x":[1,3],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([14, 28, 42, 56], 1/3, 2, [[0, 14, 14], [1, 28, 28/3], [2, 42, 14/3]], 28, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([14, 28, 42, 56], 1/3, 2, [[0, 14, 14], [1, 28, 28/3], [2, 42, 14/3]], 28, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

40. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[15,30,45,60,75],"x":[1,4],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

41. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[16,32,48,64,80,96],"x":[1,2],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

42. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[17,34,51],"x":[1,3],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([17, 34, 51], 1/3, 2, [[0, 17, 17], [1, 34, 34/3], [2, 51, 17/3]], 34, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([17, 34, 51], 1/3, 2, [[0, 17, 17], [1, 34, 34/3], [2, 51, 17/3]], 34, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

43. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[18,36,54,72],"x":[1,4],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([18, 36, 54, 72], 1/4, 2, [[0, 18, 18], [1, 36, 9], [2, 54, 27/8]], 243/8, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([18, 36, 54, 72], 1/4, 2, [[0, 18, 18], [1, 36, 9], [2, 54, 27/8]], 243/8, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

44. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[19,38,57,76,95],"x":[1,2],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

45. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[20,40,60,80,100,120],"x":[1,3],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

46. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[21,42,63],"x":[1,4],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([21, 42, 63], 1/4, 2, [[0, 21, 21], [1, 42, 21/2], [2, 63, 63/16]], 567/16, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([21, 42, 63], 1/4, 2, [[0, 21, 21], [1, 42, 21/2], [2, 63, 63/16]], 567/16, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

47. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[22,44,66,88],"x":[1,2],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([22, 44, 66, 88], 1/2, 2, [[0, 22, 22], [1, 44, 22], [2, 66, 33/2]], 121/2, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([22, 44, 66, 88], 1/2, 2, [[0, 22, 22], [1, 44, 22], [2, 66, 33/2]], 121/2, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

48. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[23,46,69,92,115],"x":[1,3],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

49. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[24,48,72,96,120,144],"x":[1,4],"cutoff":4}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([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, 否, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

50. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-finite-partial-sum;資料={"finite_support_coefficients":[25,50,75],"x":[1,2],"cutoff":2}。請依 [finite_support_coefficients, x, cutoff, term_rows, finite_sum, complete_support_included, certificate] 欄序回答。

complete_support_included核完整非零support是否已包含,不是單一x處的偶然值相等測試;未全包含時在x=0仍可能數值相等。

查看答案與解釋

([25, 50, 75], 1/2, 2, [[0, 25, 25], [1, 50, 25], [2, 75, 75/4]], 275/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included)

Wave6第七層核對=([25, 50, 75], 1/2, 2, [[0, 25, 25], [1, 50, 25], [2, 75, 75/4]], 275/4, 是, finite_partial_sum_equals_the_full_OGF_only_after_all_nonzero_finite_support_coefficients_are_included);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-finite-partial-sum · candidate

51. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[1,2],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(1/2)^n_for_every_n>=0, 1/2, 1/4, 4/3, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(1/2)^n_for_every_n>=0, 1/2, 1/4, 4/3, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

52. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[2,3],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(2/3)^n_for_every_n>=0, 1/2, 1/3, 3/2, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(2/3)^n_for_every_n>=0, 1/2, 1/3, 3/2, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

53. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[3,4],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(3/4)^n_for_every_n>=0, 1/2, 3/8, 8/5, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(3/4)^n_for_every_n>=0, 1/2, 3/8, 8/5, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

54. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[4,5],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(4/5)^n_for_every_n>=0, 1/2, 2/5, 5/3, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(4/5)^n_for_every_n>=0, 1/2, 2/5, 5/3, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

55. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[5,6],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(5/6)^n_for_every_n>=0, 1/2, 5/12, 12/7, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(5/6)^n_for_every_n>=0, 1/2, 5/12, 12/7, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

56. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[6,7],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(6/7)^n_for_every_n>=0, 1/2, 3/7, 7/4, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(6/7)^n_for_every_n>=0, 1/2, 3/7, 7/4, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

57. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[7,8],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(7/8)^n_for_every_n>=0, 1/2, 7/16, 16/9, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(7/8)^n_for_every_n>=0, 1/2, 7/16, 16/9, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

58. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[8,9],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(8/9)^n_for_every_n>=0, 1/2, 4/9, 9/5, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(8/9)^n_for_every_n>=0, 1/2, 4/9, 9/5, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

59. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[9,10],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(9/10)^n_for_every_n>=0, 1/2, 9/20, 20/11, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(9/10)^n_for_every_n>=0, 1/2, 9/20, 20/11, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

60. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[10,11],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(10/11)^n_for_every_n>=0, 1/2, 5/11, 11/6, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(10/11)^n_for_every_n>=0, 1/2, 5/11, 11/6, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

61. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[11,12],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(11/12)^n_for_every_n>=0, 1/2, 11/24, 24/13, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(11/12)^n_for_every_n>=0, 1/2, 11/24, 24/13, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

62. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[12,13],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(12/13)^n_for_every_n>=0, 1/2, 6/13, 13/7, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(12/13)^n_for_every_n>=0, 1/2, 6/13, 13/7, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

63. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[13,14],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(13/14)^n_for_every_n>=0, 1/2, 13/28, 28/15, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(13/14)^n_for_every_n>=0, 1/2, 13/28, 28/15, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

64. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[14,15],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(14/15)^n_for_every_n>=0, 1/2, 7/15, 15/8, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(14/15)^n_for_every_n>=0, 1/2, 7/15, 15/8, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

65. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[15,16],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(15/16)^n_for_every_n>=0, 1/2, 15/32, 32/17, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(15/16)^n_for_every_n>=0, 1/2, 15/32, 32/17, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

66. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[16,17],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(16/17)^n_for_every_n>=0, 1/2, 8/17, 17/9, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(16/17)^n_for_every_n>=0, 1/2, 8/17, 17/9, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

67. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[17,18],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(17/18)^n_for_every_n>=0, 1/2, 17/36, 36/19, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(17/18)^n_for_every_n>=0, 1/2, 17/36, 36/19, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

68. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[18,19],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(18/19)^n_for_every_n>=0, 1/2, 9/19, 19/10, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(18/19)^n_for_every_n>=0, 1/2, 9/19, 19/10, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

69. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[19,20],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(19/20)^n_for_every_n>=0, 1/2, 19/40, 40/21, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(19/20)^n_for_every_n>=0, 1/2, 19/40, 40/21, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

70. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[20,21],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(20/21)^n_for_every_n>=0, 1/2, 10/21, 21/11, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(20/21)^n_for_every_n>=0, 1/2, 10/21, 21/11, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

71. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[21,22],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(21/22)^n_for_every_n>=0, 1/2, 21/44, 44/23, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(21/22)^n_for_every_n>=0, 1/2, 21/44, 44/23, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

72. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[22,23],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(22/23)^n_for_every_n>=0, 1/2, 11/23, 23/12, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(22/23)^n_for_every_n>=0, 1/2, 11/23, 23/12, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

73. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[23,24],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(23/24)^n_for_every_n>=0, 1/2, 23/48, 48/25, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(23/24)^n_for_every_n>=0, 1/2, 23/48, 48/25, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

74. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[24,25],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(24/25)^n_for_every_n>=0, 1/2, 12/25, 25/13, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(24/25)^n_for_every_n>=0, 1/2, 12/25, 25/13, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

75. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-geometric-closed-form;資料={"ratio":[25,26],"x":[1,2]}。請依 [coefficient_law, x, ratio_times_x, analytic_value, domain_certificate, certificate] 欄序回答。

查看答案與解釋

(a_n=(25/26)^n_for_every_n>=0, 1/2, 25/52, 52/27, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one)

Wave6第七層核對=(a_n=(25/26)^n_for_every_n>=0, 1/2, 25/52, 52/27, 是, A(x)=1/(1-rx)_only_on_absolute_rx_less_than_one);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-geometric-closed-form · candidate

76. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":1,"a1":2,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

77. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":2,"a1":3,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

78. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":3,"a1":4,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

79. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":4,"a1":5,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

80. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":5,"a1":6,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

81. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":6,"a1":7,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

82. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":7,"a1":8,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

83. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":8,"a1":9,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

84. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":9,"a1":10,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

85. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":10,"a1":11,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

86. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":11,"a1":12,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

87. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":12,"a1":13,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

88. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":13,"a1":14,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

89. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":14,"a1":15,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

90. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":15,"a1":16,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

91. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":16,"a1":17,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

92. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":17,"a1":18,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

93. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":18,"a1":19,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

94. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":19,"a1":20,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

95. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":20,"a1":21,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

96. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":21,"a1":22,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

97. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":22,"a1":23,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

98. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":23,"a1":24,"x":[1,5]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

99. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":24,"a1":25,"x":[1,6]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

100. 解析普通生成函數

A(x)=sum_n>=0 a_n x^n的完整sequence law與明示analytic domain。操作=ogf-recurrence-boundary-terms;資料={"a0":25,"a1":26,"x":[1,4]}。請依 [a0, a1, coefficient_preview, x, numerator, denominator, analytic_value, certificate] 欄序回答。

查看答案與解釋

(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)

Wave6第七層核對=(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);依完整明示資料獨立重算;finite-support、geometric或二階recurrence law均完整明示;recurrence保留initial boundary terms;formal identity不自動是analytic equality;radius外代值、finite-prefix決定完整OGF與未證termwise differentiation均禁止。

本題逐步詳解(AI 二次覆核)

解析普通生成函數 · qf-mod44-ogf-recurrence-boundary-terms · candidate

原始題目 JSONL · 驗收報告