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CorollaryStatement: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-5.6-terra)audited 2026-08-28
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Bracelet count from the dihedral-group cycle index

Statement

For integers n≥1 and m≥1, the number of length-n bracelets over an m-letter alphabet is:

  • if n is odd,

    12n∑d∣nφ(d)mn/d+12m(n+1)/2;

  • if n is even,

    12n∑d∣nφ(d)mn/d+14(mn/2+mn/2+1).

Facts & Assumptions

Given: integers n≥1 and m≥1.

[L1]

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

[L2]

The dihedral cycle index is the odd/even formula of The cycle index of the dihedral group D_{2n}.

Proof

technique · direct
1.1L1construct

A bracelet is a colouring of a labelled n-gon up to all dihedral symmetries, so [L1] counts it by evaluating Z(D2n) at sd=m for all d.

2.1step 1.1L2algebra

Substitute sd=m into the odd case of [L2]. Since s1s2(n−1)/2 becomes m(n+1)/2, the odd-n bracelet count is 12n∑d∣nφ(d)mn/d+12m(n+1)/2.

2.2step 1.1L2algebra

Substitute sd=m into the even case of [L2]. The two reflection monomials become mn/2+1 and mn/2, giving 12n∑d∣nφ(d)mn/d+14(mn/2+mn/2+1).

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 are the two parity cases for n, so they prove the stated bracelet formulas.

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