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Bracelet count from the dihedral-group cycle index

Statement

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

if $n$ is odd,

$\frac{1}{2 n} \sum_{d ∣ n} φ (d) m^{n / d} + \frac{1}{2} m^{(n + 1) / 2};$
if $n$ is even,

$\frac{1}{2 n} \sum_{d ∣ n} φ (d) m^{n / d} + \frac{1}{4} (m^{n / 2} + m^{n / 2 + 1}) .$

Facts & Assumptions

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

[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

1.1

A bracelet is a colouring of a labelled $n$ -gon up to all dihedral symmetries, so [L1] counts it by evaluating $Z (D_{2 n})$ at $s_{d} = m$ for all $d$ .

L1construct

2.1

Substitute $s_{d} = m$ into the odd case of [L2]. Since $s_{1} s_{2}^{(n - 1) / 2}$ becomes $m^{(n + 1) / 2}$ , the odd- $n$ bracelet count is $\frac{1}{2 n} \sum_{d ∣ n} φ (d) m^{n / d} + \frac{1}{2} m^{(n + 1) / 2}$ .

step 1.1L2algebra

2.2

Substitute $s_{d} = m$ into the even case of [L2]. The two reflection monomials become $m^{n / 2 + 1}$ and $m^{n / 2}$ , giving $\frac{1}{2 n} \sum_{d ∣ n} φ (d) m^{n / d} + \frac{1}{4} (m^{n / 2} + m^{n / 2 + 1})$ .

step 1.1L2algebra

3.1

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

step 2.1step 2.2∎

Depends on

Used by

Two-colour bracelets of length 6 Example

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)