PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

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The divisor functions arise by Dirichlet convolution

Statement

One has

$τ = 1 * 1, σ_{k} = 1 * {id}_{k} .$

Consequently $τ$ and every $σ_{k}$ are multiplicative. For a prime power $p^{a}$ ,

$τ (p^{a}) = a + 1, σ_{k} (p^{a}) = 1 + p^{k} + \dots + p^{a k} .$

Facts & Assumptions

Given: An integer $k$ , a positive integer $n$ , and a prime power $p^{a}$ with $a \geq 0$ .

Proof

technique · direct

1.1givenalgebra

By Dirichlet convolution of arithmetic functions, one has $(1 * 1) (n) = \sum_{d ∣ n} 1 = τ (n)$ from The divisor-counting function $τ$ , and also $(1 * {id}_{k}) (n) = \sum_{d ∣ n} d^{k} = σ_{k} (n)$ from The power functions ${id}_{k}$ and the divisor-power-sum functions $σ_{k}$ .

2.1step 1.1given

The functions $1$ and ${id}_{k}$ are completely multiplicative in the sense of Completely multiplicative arithmetic functions, so Dirichlet convolution preserves multiplicativity, and multiplicative inverses stay multiplicative makes $τ$ and $σ_{k}$ multiplicative.

3.1givenalgebra∎

The positive divisors of $p^{a}$ are exactly $1, p, \dots, p^{a}$ . Therefore $τ (p^{a}) = \sum_{j = 0}^{a} 1 = a + 1$ and $σ_{k} (p^{a}) = \sum_{j = 0}^{a} p^{j k} = 1 + p^{k} + \dots + p^{a k}$ .

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kiran S. Kedlaya, An Introduction to Analytic Number Theory, Definition 3.8 (standard reference, not scraped)
Victor Shoup, A Computational Introduction to Number Theory and Algebra, Exercises 2.49 and 2.50 (standard reference, not scraped)