第 21705 題：Möbius函數

原題與已知條件

positive integer domain與完整UPF multiplicities。操作=mobius-canonical-value；資料={"n":30}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — positive integer domain與完整UPF multiplicities。操作=mobius-canonical-value；資料={"n":30}。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-mobius-function

逐步推導

source operation=mobius-canonical-value;original snapshot and mathematics retained

explicit ordered source output fields=[n, prime_multiset, distinct_primes, mu, certificate]

mobius original n:actual positive input for mu=30

complete prime multiplicities required for this input;mu(1) uses empty product,not a partial factor search

ORIGINAL positive integer absolute magnitude=30>1

initialize remaining=n,prime multiset empty;invariant product(extracted)*remaining=n;every smaller trial divisor has been exhausted

upf extract divisor2 occurrence0 := div(30, 2)

upf extract divisor2 occurrence0 = 15

next divisor=2;previous remaining=30;all smaller divisors exhausted

extracted multiplicity=1;next remaining=15;remaining modulo2=1;extracted multiset=[2]

upf extract divisor3 occurrence1 := div(15, 3)

upf extract divisor3 occurrence1 = 5

next divisor=3;previous remaining=15;all smaller divisors exhausted

extracted multiplicity=1;next remaining=5;remaining modulo3=2;extracted multiset=[2, 3]

upf canonical complete product factor0 := mul(1, 2)

upf canonical complete product factor0 = 2

upf canonical complete product factor1 := mul(2, 3)

upf canonical complete product factor1 = 6

upf canonical complete product factor2 := mul(6, 5)

upf canonical complete product factor2 = 30

exhausted trial divisors below4;terminal remaining=5;d*d>remaining;ORIGINAL n=30

complete increasing prime multiset=[2, 3, 5];full product=30;a composite terminal remainder would have an already tested smaller divisor

mobius original n:input=30;COMPLETE prime multiset=[2, 3, 5];distinct primes=[2, 3, 5]

mu=-1;empty product case=否;repeated prime=否

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 upf canonical complete product factor2，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 upf canonical complete product factor1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 upf canonical complete product factor0，不以小數近似替代。
• S007 · 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。
• S006 · premise_and_result_audit — 以精確有理數重算 upf extract divisor3 occurrence1，不以小數近似替代。
• S005 · 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。
• S004 · premise_and_result_audit — 以精確有理數重算 upf extract divisor2 occurrence0，不以小數近似替代。
• S003 · 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。
• 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

(30, [2, 3, 5], [2, 3, 5], -1, squarefree_parity_sign)

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。