Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31
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The divisor functions arise by Dirichlet convolution

Statement

One has

τ=11,σk=1idk.

Consequently τ and every σk are multiplicative. For a prime power pa,

τ(pa)=a+1,σk(pa)=1+pk++pak.

Facts & Assumptions

Given: An integer k, a positive integer n, and a prime power pa with a0.

Proof

technique · direct
1.1

By Dirichlet convolution of arithmetic functions, one has (11)(n)=dn1=τ(n) from The divisor-counting function τ, and also (1idk)(n)=dndk=σk(n) from The power functions idk and the divisor-power-sum functions σk.

givenalgebra
2.1

The functions 1 and idk are completely multiplicative in the sense of Completely multiplicative arithmetic functions, so Dirichlet convolution preserves multiplicativity, and multiplicative inverses stay multiplicative makes τ and σk multiplicative.

step 1.1given
3.1

The positive divisors of pa are exactly 1,p,,pa. Therefore τ(pa)=j=0a1=a+1 and σk(pa)=j=0apjk=1+pk++pak.

givenalgebra

Depends on

Used by

Dependency tree · two levels

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Sources