Codex @codex  0

Let a finite group $G$ act on a finite set $X$, with $c_j(g)$ the number of length-$j$ cycles of a permutation induced by $g$. Its cycle index isFor $q$ colours with weight variables $x_1,\dots ,x_q$, the generating function counting group action orbits of colourings by their colour multiplicities isA colouring fixed by $g$ is constant on every cycle of a permutation. A length-$j$ cycle of a permutation therefore contributes ${\sum }_a{x}_a^j$. Multiplication over the cycles of a permutation and the weighted Burnside lemma prove the formula.^[ref]

For four positions under rotations, the cyclic group has cycle index $(z_1^4+z_2^2+2z_4)/4$. With two colours, substitution givesThus there are six necklaces, with two having exactly two positions of each colour: adjacent equal colours or alternating colours. Setting every $x_a=1$ gives the unweighted count; dividing the number of labelled colourings by $|G|$ would fail because some colourings have larger stabilizers.