第 22593 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":[2,4,6,8],"constant":0,"initial_values":[17,18,19,20]}。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":[2,4,6,8],"constant":0,"initial_values":[17,18,19,20]}。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=[2, 4, 6, 8];constant=0;ALL initial a_0..a_(r-1)=[17, 18, 19, 20]

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(2, 17)

recurrence ORIGINAL n=4 ORIGINAL coefficient0 times a_0 = 34

recurrence ORIGINAL n=4 partial sum after coefficient0 := sum(0, 34)

recurrence ORIGINAL n=4 partial sum after coefficient0 = 34

recurrence ORIGINAL n=4 ORIGINAL coefficient1 times a_1 := mul(4, 18)

recurrence ORIGINAL n=4 ORIGINAL coefficient1 times a_1 = 72

recurrence ORIGINAL n=4 partial sum after coefficient1 := sum(34, 72)

recurrence ORIGINAL n=4 partial sum after coefficient1 = 106

recurrence ORIGINAL n=4 ORIGINAL coefficient2 times a_2 := mul(6, 19)

recurrence ORIGINAL n=4 ORIGINAL coefficient2 times a_2 = 114

recurrence ORIGINAL n=4 partial sum after coefficient2 := sum(106, 114)

recurrence ORIGINAL n=4 partial sum after coefficient2 = 220

recurrence ORIGINAL n=4 ORIGINAL coefficient3 times a_3 := mul(8, 20)

recurrence ORIGINAL n=4 ORIGINAL coefficient3 times a_3 = 160

recurrence ORIGINAL n=4 partial sum after coefficient3 := sum(220, 160)

recurrence ORIGINAL n=4 partial sum after coefficient3 = 380

recurrence ORIGINAL:ORIGINAL target n=4;previous state a_(n-r)..a_(n-1)=[17, 18, 19, 20];ALL [j,source_index,coefficient,value,product] rows=[[0, 0, 2, 17, 34], [1, 1, 4, 18, 72], [2, 2, 6, 19, 114], [3, 3, 8, 20, 160]]

new a_4=380;next state=[18, 19, 20, 380];drop ONLY oldest state entry then append new scalar;prefix now=[17, 18, 19, 20, 380]

recurrence ORIGINAL:complete ORIGINAL initial state=[17, 18, 19, 20];actual generated prefix=[17, 18, 19, 20, 380]

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, [2, 4, 6, 8], 0, [17, 18, 19, 20], 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。