Catalan數基礎 · 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-20261111-r05-catalan-numbers-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. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":0}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(0, [1], 1, C_zero_equals_one)
Wave6第五層核對=(0, [1], 1, C_zero_equals_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
2. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":1}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(1, [1, 1], 1, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(1, [1, 1], 1, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
3. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":2}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(2, [1, 1, 2], 2, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(2, [1, 1, 2], 2, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
4. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":3}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(3, [1, 1, 2, 5], 5, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(3, [1, 1, 2, 5], 5, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
5. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":4}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(4, [1, 1, 2, 5, 14], 14, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(4, [1, 1, 2, 5, 14], 14, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
6. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":5}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(5, [1, 1, 2, 5, 14, 42], 42, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(5, [1, 1, 2, 5, 14, 42], 42, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
7. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":6}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(6, [1, 1, 2, 5, 14, 42, 132], 132, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(6, [1, 1, 2, 5, 14, 42, 132], 132, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
8. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":7}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(7, [1, 1, 2, 5, 14, 42, 132, 429], 429, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(7, [1, 1, 2, 5, 14, 42, 132, 429], 429, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
9. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":8}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(8, [1, 1, 2, 5, 14, 42, 132, 429, 1430], 1430, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(8, [1, 1, 2, 5, 14, 42, 132, 429, 1430], 1430, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
10. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":9}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(9, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862], 4862, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(9, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862], 4862, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
11. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":10}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(10, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796], 16796, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(10, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796], 16796, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
12. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":11}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(11, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786], 58786, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(11, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786], 58786, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
13. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":12}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(12, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012], 208012, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(12, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012], 208012, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
14. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":13}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(13, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900], 742900, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(13, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900], 742900, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
15. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":14}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(14, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440], 2674440, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(14, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440], 2674440, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
16. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":15}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(15, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845], 9694845, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(15, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845], 9694845, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
17. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":16}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(16, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670], 35357670, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(16, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670], 35357670, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
18. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":17}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(17, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790], 129644790, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(17, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790], 129644790, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
19. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":18}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(18, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700], 477638700, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(18, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700], 477638700, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
20. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":19}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(19, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190], 1767263190, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(19, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190], 1767263190, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
21. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":20}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(20, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420], 6564120420, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(20, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420], 6564120420, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
22. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":21}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(21, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020], 24466267020, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(21, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020], 24466267020, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
23. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":22}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(22, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640], 91482563640, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(22, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640], 91482563640, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
24. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":23}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(23, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650], 343059613650, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(23, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650], 343059613650, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
25. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-generation;資料={"n":24}。請依 [n, prefix, C_n, index_convention] 欄序回答。
查看答案與解釋
(24, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324], 1289904147324, C_next_sum_i_zero_to_n_minus_one)
Wave6第五層核對=(24, [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324], 1289904147324, C_next_sum_i_zero_to_n_minus_one);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-generation · candidate
26. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":0}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(0, 1, 1, 0, 1, 1, 是)
Wave6第五層核對=(0, 1, 1, 0, 1, 1, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
27. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":1}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(1, 2, 2, 0, 1, 1, 是)
Wave6第五層核對=(1, 2, 2, 0, 1, 1, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
28. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":2}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(2, 6, 3, 0, 2, 2, 是)
Wave6第五層核對=(2, 6, 3, 0, 2, 2, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
29. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":3}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(3, 20, 4, 0, 5, 5, 是)
Wave6第五層核對=(3, 20, 4, 0, 5, 5, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
30. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":4}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(4, 70, 5, 0, 14, 14, 是)
Wave6第五層核對=(4, 70, 5, 0, 14, 14, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
31. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":5}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(5, 252, 6, 0, 42, 42, 是)
Wave6第五層核對=(5, 252, 6, 0, 42, 42, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
32. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":6}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(6, 924, 7, 0, 132, 132, 是)
Wave6第五層核對=(6, 924, 7, 0, 132, 132, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
33. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":7}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(7, 3432, 8, 0, 429, 429, 是)
Wave6第五層核對=(7, 3432, 8, 0, 429, 429, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
34. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":8}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(8, 12870, 9, 0, 1430, 1430, 是)
Wave6第五層核對=(8, 12870, 9, 0, 1430, 1430, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
35. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":9}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(9, 48620, 10, 0, 4862, 4862, 是)
Wave6第五層核對=(9, 48620, 10, 0, 4862, 4862, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
36. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":10}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(10, 184756, 11, 0, 16796, 16796, 是)
Wave6第五層核對=(10, 184756, 11, 0, 16796, 16796, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
37. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":11}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(11, 705432, 12, 0, 58786, 58786, 是)
Wave6第五層核對=(11, 705432, 12, 0, 58786, 58786, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
38. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":12}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(12, 2704156, 13, 0, 208012, 208012, 是)
Wave6第五層核對=(12, 2704156, 13, 0, 208012, 208012, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
39. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":13}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(13, 10400600, 14, 0, 742900, 742900, 是)
Wave6第五層核對=(13, 10400600, 14, 0, 742900, 742900, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
40. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":14}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(14, 40116600, 15, 0, 2674440, 2674440, 是)
Wave6第五層核對=(14, 40116600, 15, 0, 2674440, 2674440, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
41. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":15}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(15, 155117520, 16, 0, 9694845, 9694845, 是)
Wave6第五層核對=(15, 155117520, 16, 0, 9694845, 9694845, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
42. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":16}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(16, 601080390, 17, 0, 35357670, 35357670, 是)
Wave6第五層核對=(16, 601080390, 17, 0, 35357670, 35357670, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
43. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":17}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(17, 2333606220, 18, 0, 129644790, 129644790, 是)
Wave6第五層核對=(17, 2333606220, 18, 0, 129644790, 129644790, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
44. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":18}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(18, 9075135300, 19, 0, 477638700, 477638700, 是)
Wave6第五層核對=(18, 9075135300, 19, 0, 477638700, 477638700, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
45. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":19}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(19, 35345263800, 20, 0, 1767263190, 1767263190, 是)
Wave6第五層核對=(19, 35345263800, 20, 0, 1767263190, 1767263190, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
46. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":20}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(20, 137846528820, 21, 0, 6564120420, 6564120420, 是)
Wave6第五層核對=(20, 137846528820, 21, 0, 6564120420, 6564120420, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
47. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":21}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(21, 538257874440, 22, 0, 24466267020, 24466267020, 是)
Wave6第五層核對=(21, 538257874440, 22, 0, 24466267020, 24466267020, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
48. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":22}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(22, 2104098963720, 23, 0, 91482563640, 91482563640, 是)
Wave6第五層核對=(22, 2104098963720, 23, 0, 91482563640, 91482563640, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
49. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":23}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(23, 8233430727600, 24, 0, 343059613650, 343059613650, 是)
Wave6第五層核對=(23, 8233430727600, 24, 0, 343059613650, 343059613650, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
50. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-closed-form;資料={"n":24}。請依 [n, central_binomial, denominator, remainder, quotient, binomial_difference, equal] 欄序回答。
查看答案與解釋
(24, 32247603683100, 25, 0, 1289904147324, 1289904147324, 是)
Wave6第五層核對=(24, 32247603683100, 25, 0, 1289904147324, 1289904147324, 是);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-closed-form · candidate
51. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":0}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(0, 1, 1, 1, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(0, 1, 1, 1, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
52. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":1}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(1, 1, 1, 1, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(1, 1, 1, 1, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
53. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":2}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(2, 2, 2, 2, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(2, 2, 2, 2, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
54. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":3}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(3, 5, 5, 5, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(3, 5, 5, 5, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
55. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":4}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(4, 14, 14, 14, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(4, 14, 14, 14, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
56. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":5}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(5, 42, 42, 42, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(5, 42, 42, 42, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
57. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":6}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(6, 132, 132, 132, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(6, 132, 132, 132, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
58. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":7}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(7, 429, 429, 429, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(7, 429, 429, 429, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
59. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":8}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(8, 1430, 1430, 1430, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(8, 1430, 1430, 1430, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
60. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":9}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(9, 4862, 4862, 4862, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(9, 4862, 4862, 4862, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
61. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":10}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(10, 16796, 16796, 16796, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(10, 16796, 16796, 16796, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
62. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":11}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(11, 58786, 58786, 58786, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(11, 58786, 58786, 58786, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
63. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":12}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(12, 208012, 208012, 208012, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(12, 208012, 208012, 208012, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
64. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":13}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(13, 742900, 742900, 742900, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(13, 742900, 742900, 742900, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
65. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":14}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(14, 2674440, 2674440, 2674440, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(14, 2674440, 2674440, 2674440, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
66. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":15}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(15, 9694845, 9694845, 9694845, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(15, 9694845, 9694845, 9694845, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
67. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":16}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(16, 35357670, 35357670, 35357670, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(16, 35357670, 35357670, 35357670, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
68. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":17}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(17, 129644790, 129644790, 129644790, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(17, 129644790, 129644790, 129644790, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
69. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":18}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(18, 477638700, 477638700, 477638700, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(18, 477638700, 477638700, 477638700, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
70. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":19}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(19, 1767263190, 1767263190, 1767263190, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(19, 1767263190, 1767263190, 1767263190, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
71. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":20}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(20, 6564120420, 6564120420, 6564120420, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(20, 6564120420, 6564120420, 6564120420, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
72. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":21}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(21, 24466267020, 24466267020, 24466267020, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(21, 24466267020, 24466267020, 24466267020, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
73. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":22}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(22, 91482563640, 91482563640, 91482563640, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(22, 91482563640, 91482563640, 91482563640, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
74. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":23}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(23, 343059613650, 343059613650, 343059613650, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(23, 343059613650, 343059613650, 343059613650, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
75. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-recurrence-closed-cross-check;資料={"n":24}。請依 [n, recurrence_value, closed_value, difference_value, all_equal, certificate] 欄序回答。
查看答案與解釋
(24, 1289904147324, 1289904147324, 1289904147324, 是, C_zero_index_convention_preserved)
Wave6第五層核對=(24, 1289904147324, 1289904147324, 1289904147324, 是, C_zero_index_convention_preserved);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-recurrence-closed-cross-check · candidate
76. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-1}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-1, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-1, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
77. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-2}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-2, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-2, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
78. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-3}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-3, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-3, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
79. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-4}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-4, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-4, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
80. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-5}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-5, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-5, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
81. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-6}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-6, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-6, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
82. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-7}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-7, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-7, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
83. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-8}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-8, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-8, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
84. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-9}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-9, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-9, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
85. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-10}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-10, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-10, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
86. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-11}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-11, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-11, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
87. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-12}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-12, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-12, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
88. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-13}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-13, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-13, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
89. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-14}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-14, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-14, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
90. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-15}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-15, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-15, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
91. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-16}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-16, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-16, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
92. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-17}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-17, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-17, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
93. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-18}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-18, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-18, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
94. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-19}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-19, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-19, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
95. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-20}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-20, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-20, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
96. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-21}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-21, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-21, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
97. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-22}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-22, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-22, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
98. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-23}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-23, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-23, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
99. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-24}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-24, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-24, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
100. Catalan數
C_0=1且C_(n+1)=sum_i C_i*C_(n-i)的zero-based indexing。操作=catalan-negative-index-boundary;資料={"n":-25}。請依 [n, in_domain, value_or_null, boundary] 欄序回答。
查看答案與解釋
(-25, 否, null, Catalan_domain_is_natural_numbers_including_zero)
Wave6第五層核對=(-25, 否, null, Catalan_domain_is_natural_numbers_including_zero);依完整明示資料獨立重算;closed form division前必須驗(n+1) divides binomial(2n,n),並與binomial difference cross-check;domain為n>=0;不可混用C_0/C_1 convention、negative index或integer truncation。
本題逐步詳解(AI 二次覆核)
Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate
原始題目 JSONL · 驗收報告