第 22779 題：Catalan數

原題與已知條件

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 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 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 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-catalan-numbers

逐步推導

source operation=catalan-recurrence-generation;original snapshot and mathematics retained

explicit ordered source output fields=[n, prefix, C_n, index_convention]

ORIGINAL requested natural index n=3

initialize C_0=1;recurrence C_(j+1)=sum_i=0..j C_i*C_(j-i),including ALL splits

Catalan complete convolution for C_1 := sum(1)

Catalan complete convolution for C_1 = 1

ORIGINAL index j=0;ALL [i,C_i,C_(j-i),exact product] rows=[[0, 1, 1, 1]]

C_1=1;ALL earlier values used from the established prefix=[1]

Catalan complete convolution for C_2 := sum(1, 1)

Catalan complete convolution for C_2 = 2

ORIGINAL index j=1;ALL [i,C_i,C_(j-i),exact product] rows=[[0, 1, 1, 1], [1, 1, 1, 1]]

C_2=2;ALL earlier values used from the established prefix=[1, 1]

Catalan complete convolution for C_3 := sum(2, 1, 2)

Catalan complete convolution for C_3 = 5

ORIGINAL index j=2;ALL [i,C_i,C_(j-i),exact product] rows=[[0, 1, 2, 2], [1, 1, 1, 1], [2, 2, 1, 2]]

C_3=5;ALL earlier values used from the established prefix=[1, 1, 2]

Catalan central:ORIGINAL binomial N=6,K=3

use k=min(K,N-K)=3;initialize exact product1;empty product for k0 is1

Catalan central binomial product at i=1 := mul(1, 4)

Catalan central binomial product at i=1 = 4

Catalan central exact binomial division by i=1 := div(4, 1)

Catalan central exact binomial division by i=1 = 4

Catalan central binomial product at i=2 := mul(4, 5)

Catalan central binomial product at i=2 = 20

Catalan central exact binomial division by i=2 := div(20, 2)

Catalan central exact binomial division by i=2 = 10

Catalan central binomial product at i=3 := mul(10, 6)

Catalan central binomial product at i=3 = 60

Catalan central exact binomial division by i=3 := div(60, 3)

Catalan central exact binomial division by i=3 = 20

Catalan ORIGINAL denominator n plus1 := sum(3, 1)

Catalan ORIGINAL denominator n plus1 = 4

ORIGINAL n=3;central binomial=20;positive denominator n+1=4

Euclidean quotient=5,remainder=0;check remainder0 BEFORE treating quotient as Catalan integer

Catalan exact closed quotient after remainder0 := div(20, 4)

Catalan exact closed quotient after remainder0 = 5

Catalan adjacent:ORIGINAL binomial N=6,K=4

use k=min(K,N-K)=2;initialize exact product1;empty product for k0 is1

Catalan adjacent binomial product at i=1 := mul(1, 5)

Catalan adjacent binomial product at i=1 = 5

Catalan adjacent exact binomial division by i=1 := div(5, 1)

Catalan adjacent exact binomial division by i=1 = 5

Catalan adjacent binomial product at i=2 := mul(5, 6)

Catalan adjacent binomial product at i=2 = 30

Catalan adjacent exact binomial division by i=2 := div(30, 2)

Catalan adjacent exact binomial division by i=2 = 15

Catalan central minus adjacent binomial := sub(20, 15)

Catalan central minus adjacent binomial = 5

ORIGINAL n=3;recurrence value=5;central=20,adjacent=15,denominator=4

closed quotient=5;binomial difference=5;ALL THREE agree;balanced paths counted by choose(2n,n),bad paths first-crossing reflection counted by choose(2n,n+1)

逐步反向核對

• S025 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S024 · premise_and_result_audit — 以精確有理數重算 Catalan central minus adjacent binomial，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Catalan adjacent exact binomial division by i=2，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Catalan adjacent binomial product at i=2，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Catalan adjacent exact binomial division by i=1，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Catalan adjacent binomial product at i=1，不以小數近似替代。
• S019 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S018 · premise_and_result_audit — 以精確有理數重算 Catalan exact closed quotient after remainder0，不以小數近似替代。
• S017 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S016 · premise_and_result_audit — 以精確有理數重算 Catalan ORIGINAL denominator n plus1，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Catalan central exact binomial division by i=3，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Catalan central binomial product at i=3，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Catalan central exact binomial division by i=2，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Catalan central binomial product at i=2，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 Catalan central exact binomial division by i=1，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Catalan central binomial product at i=1，不以小數近似替代。
• S009 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S007 · premise_and_result_audit — 以精確有理數重算 Catalan complete convolution for C_3，不以小數近似替代。
• S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S005 · premise_and_result_audit — 以精確有理數重算 Catalan complete convolution for C_2，不以小數近似替代。
• S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S003 · premise_and_result_audit — 以精確有理數重算 Catalan complete convolution for C_1，不以小數近似替代。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(3, [1, 1, 2, 5], 5, C_next_sum_i_zero_to_n_minus_one)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。