第 22580 題：遞迴關係

原題與已知條件

initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":2,"initial_values":[4,5,6]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — initial_values[j]=a_j且coefficients[j]乘a_(n-r+j)的明示order-r state update。操作=recurrence-specification-qualification；資料={"order":3,"coefficients":[1,2,3],"constant":2,"initial_values":[4,5,6]}。請依 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-recurrence-relation

逐步推導

source operation=recurrence-specification-qualification;original snapshot and mathematics retained

explicit ordered source output fields=[order, coefficients, constant, initial_values, complete_initials, total_rule, index_scope]

recurrence ORIGINAL:ORIGINAL order=3;oldest-to-newest coefficients=[1, 2, 3];constant=2;ALL initial a_0..a_(r-1)=[4, 5, 6]

for EVERY n>=3,a_n=constant+sum_j coefficients[j]*a_(n-r+j);current prefix length=3;forward rule is total and single-valued

recurrence ORIGINAL n=3 ORIGINAL coefficient0 times a_0 := mul(1, 4)

recurrence ORIGINAL n=3 ORIGINAL coefficient0 times a_0 = 4

recurrence ORIGINAL n=3 partial sum after coefficient0 := sum(2, 4)

recurrence ORIGINAL n=3 partial sum after coefficient0 = 6

recurrence ORIGINAL n=3 ORIGINAL coefficient1 times a_1 := mul(2, 5)

recurrence ORIGINAL n=3 ORIGINAL coefficient1 times a_1 = 10

recurrence ORIGINAL n=3 partial sum after coefficient1 := sum(6, 10)

recurrence ORIGINAL n=3 partial sum after coefficient1 = 16

recurrence ORIGINAL n=3 ORIGINAL coefficient2 times a_2 := mul(3, 6)

recurrence ORIGINAL n=3 ORIGINAL coefficient2 times a_2 = 18

recurrence ORIGINAL n=3 partial sum after coefficient2 := sum(16, 18)

recurrence ORIGINAL n=3 partial sum after coefficient2 = 34

recurrence ORIGINAL:ORIGINAL target n=3;previous state a_(n-r)..a_(n-1)=[4, 5, 6];ALL [j,source_index,coefficient,value,product] rows=[[0, 0, 1, 4, 4], [1, 1, 2, 5, 10], [2, 2, 3, 6, 18]]

new a_3=34;next state=[5, 6, 34];drop ONLY oldest state entry then append new scalar;prefix now=[4, 5, 6, 34]

recurrence ORIGINAL:complete ORIGINAL initial state=[4, 5, 6];actual generated prefix=[4, 5, 6, 34]

ANY two sequences with SAME r initials and SAME total rule agree at n=3 by substitution,then at every later n by induction;this determines the requested prefix only

逐步反向核對

• S010 · 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。
• 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 — 以精確有理數重算 recurrence ORIGINAL n=3 partial sum after coefficient2，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=3 ORIGINAL coefficient2 times a_2，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=3 partial sum after coefficient1，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=3 ORIGINAL coefficient1 times a_1，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=3 partial sum after coefficient0，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=3 ORIGINAL coefficient0 times a_0，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, 2, 3], 2, [4, 5, 6], 是, 是, rule_applies_for_n_ge_3)

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。