第 22584 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":4,"coefficients":[1,2,3,4],"constant":1,"initial_values":[8,9,10,11]}。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":4,"coefficients":[1,2,3,4],"constant":1,"initial_values":[8,9,10,11]}。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=4;oldest-to-newest coefficients=[1, 2, 3, 4];constant=1;ALL initial a_0..a_(r-1)=[8, 9, 10, 11]

for EVERY n>=4,a_n=constant+sum_j coefficients[j]*a_(n-r+j);current prefix length=4;forward rule is total and single-valued

recurrence ORIGINAL n=4 ORIGINAL coefficient0 times a_0 := mul(1, 8)

recurrence ORIGINAL n=4 ORIGINAL coefficient0 times a_0 = 8

recurrence ORIGINAL n=4 partial sum after coefficient0 := sum(1, 8)

recurrence ORIGINAL n=4 partial sum after coefficient0 = 9

recurrence ORIGINAL n=4 ORIGINAL coefficient1 times a_1 := mul(2, 9)

recurrence ORIGINAL n=4 ORIGINAL coefficient1 times a_1 = 18

recurrence ORIGINAL n=4 partial sum after coefficient1 := sum(9, 18)

recurrence ORIGINAL n=4 partial sum after coefficient1 = 27

recurrence ORIGINAL n=4 ORIGINAL coefficient2 times a_2 := mul(3, 10)

recurrence ORIGINAL n=4 ORIGINAL coefficient2 times a_2 = 30

recurrence ORIGINAL n=4 partial sum after coefficient2 := sum(27, 30)

recurrence ORIGINAL n=4 partial sum after coefficient2 = 57

recurrence ORIGINAL n=4 ORIGINAL coefficient3 times a_3 := mul(4, 11)

recurrence ORIGINAL n=4 ORIGINAL coefficient3 times a_3 = 44

recurrence ORIGINAL n=4 partial sum after coefficient3 := sum(57, 44)

recurrence ORIGINAL n=4 partial sum after coefficient3 = 101

recurrence ORIGINAL:ORIGINAL target n=4;previous state a_(n-r)..a_(n-1)=[8, 9, 10, 11];ALL [j,source_index,coefficient,value,product] rows=[[0, 0, 1, 8, 8], [1, 1, 2, 9, 18], [2, 2, 3, 10, 30], [3, 3, 4, 11, 44]]

new a_4=101;next state=[9, 10, 11, 101];drop ONLY oldest state entry then append new scalar;prefix now=[8, 9, 10, 11, 101]

recurrence ORIGINAL:complete ORIGINAL initial state=[8, 9, 10, 11];actual generated prefix=[8, 9, 10, 11, 101]

ANY two sequences with SAME r initials and SAME total rule agree at n=4 by substitution,then at every later n by induction;this determines the requested prefix only

逐步反向核對

• S012 · 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。
• S011 · 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。
• S010 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=4 partial sum after coefficient3，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=4 ORIGINAL coefficient3 times a_3，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 recurrence ORIGINAL n=4 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=4 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=4 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=4 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=4 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=4 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

(4, [1, 2, 3, 4], 1, [8, 9, 10, 11], 是, 是, rule_applies_for_n_ge_4)

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。