# 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](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0001/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0002/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0003/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0004/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0005/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0006/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0007/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0008/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0009/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0010/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0011/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0012/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0013/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0014/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0015/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0016/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0017/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0018/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0019/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0020/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0021/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0022/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0023/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0024/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0025/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0026/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0027/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0028/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0029/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0030/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0031/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0032/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0033/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0034/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0035/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0036/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0037/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0038/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0039/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0040/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0041/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0042/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0043/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0044/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0045/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0046/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0047/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0048/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0049/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0050/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0051/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0052/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0053/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0054/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0055/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0056/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0057/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0058/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0059/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0060/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0061/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0062/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0063/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0064/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0065/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0066/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0067/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0068/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0069/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0070/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0071/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0072/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0073/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0074/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0075/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0076/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0077/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0078/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0079/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0080/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0081/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0082/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0083/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0084/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0085/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0086/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0087/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0088/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0089/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0090/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0091/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0092/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0093/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0094/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0095/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0096/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0097/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0098/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0099/)

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）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r05-catalan-numbers-0100/)

Catalan數 · qf-mod42-catalan-negative-index-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r05-catalan-numbers-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r05-catalan-numbers-foundations/validation-report.json)