CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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The number of necklaces of length $n$ on an $m$ -letter alphabet is $\frac{1}{n} \sum_{d ∣ n} φ (d) m^{n / d}$

Statement

Let $m \geq 1$ and $n \geq 1$ . The number of necklaces of length $n$ on an $m$ -letter alphabet is

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

Facts & Assumptions

Given: Naturals $m \geq 1$ and $n \geq 1$ , and the class $A : = m Z$ of $m$ coloured atoms.

[L1]

If a combinatorial class $A$ has no size-zero objects, then over a commutative $Q$ -algebra its cycle construction has generating function $OGF (CYC (A)) = \sum_{k \geq 1} \frac{φ (k)}{k} \log \frac{1}{1 - A (x^{k})}$ (Over a commutative $Q$ -algebra, $CYC (A)$ has generating function $\sum_{k \geq 1} \frac{φ (k)}{k} \log \frac{1}{1 - A (x^{k})}$ ).

Proof

technique · direct

1.1construct

The class $A$ has generating function $A (x) = m x$ . Its cycle class is exactly the class of coloured necklaces, with size equal to necklace length.

1.2algebra

For each $k \geq 1$ , one has $\log (1 / (1 - m x^{k})) = \sum_{j \geq 1} m^{j} x^{k j} / j$ , so the coefficient of $x^{n}$ in this series is $0$ unless $k ∣ n$ , and is $m^{n / k} / (n / k)$ when $k ∣ n$ .

2.1step 1.1step 1.2L1algebra∎

Taking the coefficient of $x^{n}$ in [L1] and using steps 1.1 and 1.2 gives $[x^{n}] OGF (CYC (A)) = \sum_{k ∣ n} (φ (k) / k) \cdot (m^{n / k} / (n / k)) = (1 / n) \sum_{k ∣ n} φ (k) m^{n / k}$ . This coefficient is exactly the number of necklaces of length $n$ .

Depends on

Used by

Dependency tree · two levels

10 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

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)
Stephen Melczer, An Invitation to Enumeration, Chapter 5: Combinatorial Constructions (standard reference, not scraped)

The number of necklaces of length nn on an mm-letter alphabet is 1n∑d∣nφ(d)mn/dn1​∑d∣n​φ(d)mn/d