I have a circle divided into 60 pieces, and I have 4 different colors $(c_1,c_2,c_3,c_4)$, and I want to know how many different "colorings" I can get.
So I thought I would use the Polya Enumeration theorem, and I want to find the coefficient of the term in the polynomial with $c_1^{15} c_2^{15} c_3^{15} c_4^{15} $.
For this I looked at the Dihedral Group $D_{60}$, which is generated by $<\tau,\sigma >$, where $\sigma=(1\ 2\ \ldots \ 60)$ and $\tau=(1\ \ 60)(2\ \ 59)\ldots (30\ \ 31)$(a reflection). I already found all the $\sigma^i\ , \forall i=1,\ldots,60$, I mean their representation in cycles, but I still have to find at least $\tau \sigma = ?$ and I believe maybe also $\tau^2\sigma$ and others maybe.
After that and with the information I now have I would have to calculate the polynom $P(x_1,\ldots,x_{60})=\frac{1}{|D_{60}|}(x_1^{60}+x_{60}^j+\ldots)$.
So where I would need help is that I don't know exactly which combinations of $\tau^i\sigma^j$ I have to compose. And even better, if there is an easier solution for this, because the process of calculating all the $\sigma^i$ is not short, and I would still have to compute a massive Polynom...
I believe now that I have to see the cyclic structure of $\tau \sigma^i$, and not $\tau^j\sigma^i$. But still, if someone has a shorter method I would be very happy!
To solve this by PET we require the cycle index $Z(D_{60})$ (dihedral group). We have the cycle index of the cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
Continuing we have the cycle index of the dihedral group
$$Z(D_n) = \frac{1}{2} Z(C_n) + \begin{cases} \frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\ \frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right) & n \text{ even.} \end{cases}$$
In the present case we find
$$Z(D_{60}) = {\frac {{a_{{1}}}^{60}}{120}}+1/4\,{a_{{1}}}^{2}{a_{{2}}}^{29} +{\frac {31\,{a_{{2}}}^{30}}{120}}+{\frac {{a_{{3}}}^{20}}{60}} +{\frac {{a_{{4}}}^{15}}{60}}+1/30\,{a_{{5}}}^{12} +{\frac {{a_{{6}}}^{10}}{60}}+1/30\,{a_{{10}}}^{6} \\+1/30\,{a_{{12}}}^{5}+1/15\,{a_{{15}}}^{4} +1/15\,{a_{{20}}}^{3}+1/15\,{a_{{30}}}^{2}+2/15\,a_{{60}}.$$
Working through these term by term and skipping the ones that do not contribute leaves just
The second term is special:
$$ \frac{a_1^2 a_2^{29}}{4} :[C_1^{15} C_2^{15} C_3^{15} C_4^{15}] \frac{1}{4} (C_1 + C_2 + C_3 + C_4)^{2} (C_1^2 + C_2^2 + C_3^2 + C_4^2)^{29} = 0.$$
Adding all contributions we find
$$\bbox[5px,border:2px solid #00A000]{ 23713472717216429668046350036152.}$$
As for the terms that do not contribute this is decided by examining powers. E.g. the term $\frac{1}{30} a_{10}^6$ yields $\frac{1}{30} (C_1^{10} + C_2^{10} + C_3^{10} + C_4^{10})^{6}$ (tenth powers) and as ten does not divide fifteen we cannot possibly obtain the desired distribution of colors, for a contribution of zero.