CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)judge pass (gpt-5.6-terra)audited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Necklace count from the cyclic-group cycle index

Statement

For integers $n \geq 1$ and $m \geq 1$ , the number of length- $n$ necklaces over an $m$ -letter alphabet is

$\frac{1}{n} \sum_{d ∣ n} φ (d) m^{n / d} .$

Facts & Assumptions

Given: integers $n \geq 1$ and $m \geq 1$ .

[L1]

Pólya's theorem counts colourings up to rotation by evaluating the cycle index at the number of colours (Pólya's enumeration theorem).

[L2]

The rotation action of $C_{n}$ has cycle index $Z (C_{n}) = \frac{1}{n} \sum_{d ∣ n} φ (d) s_{d}^{n / d}$ (The cycle index of the cyclic group C_n).

Proof

technique · direct

1.1

A length- $n$ necklace over an $m$ -letter alphabet is exactly a colouring of the vertices of a labelled $n$ -gon by $m$ colours, up to the rotation action of $C_{n}$ .

construct

2.1

By [L1] and [L2], the number of such orbits is $Z (C_{n}) (m, m, \dots, m) = \frac{1}{n} \sum_{d ∣ n} φ (d) m^{n / d}$ .

step 1.1L1L2∎

Depends on

Used by

Dependency tree · two levels

6 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

Ben Lynn, Polya Theory: The Cycle Index Polynomial (standard reference, not scraped)