Number of distinct red and blue colored cycles Consider a cycle graph of length $n$ where $k$ nodes are colored blue and $n-k$ are colored red. I define two vectors $x,y$ each of length $l$ where $l$ is the number of connected components of red and blue colored subgraphs and the entries in the vectors are defined by the size of said subgraphs. $x$ and $y$ always have to be of the same length $l$ since the coloring alternates between blue and red colored sub-graphs.
I want to find the number of distinct configurations and, hence, I have to get rid of those who I count twice due to symmetries implied by the symmetric group $S_l$ on the vectors $l$.
To rewrite this I consider two vectors of positive integers which I call obviously $x,y$ such that the length of $x$ is $l$ and the length of $y$ is likewise $l$ with $x_i\in\{1,...,k\}$ and $y_i\in\{1,...,n-k\}$ for $i=1,...,l$ and $\sum_{i=1}^l x_i = k$ and $\sum_{i=1}^l y_i = n-k$ for some positive integer $n>k$. Then two configurations are identical under a rotation of $2\pi \dfrac{r}{n}$ iff $x_i = x_{(i+r)\mathrm{mod}\;l}$ and $y_i = y_{(i+r)\mathrm{mod}\;l}$ since both $x$ and $y$ have to be rotated. Note that if you find such an $r$ for a pair $x,y$ it is not true that the same $r$ is preserved under arbitrary actions of $S_l$ on $x$ and $y$.
Can these numbers be counted or is there no hope to even get a recursive formula for the number of "identical" $x$ ad $y$ which can be rotated in this way?
 A: We can use the Pólya enumeration theorem to solve the problem for any particular $n,k,l$. I haven't been able to extract a formula for the general solution.
Think of the sequences $x$ and $y$ as "coloring" the set $\{1,2,\dots,l\}$ by giving element $i$ the "color" $(x_i, y_i)$. If we give color $(x_i, y_i)$ weight $(x_i, y_i)$ as well, then the condition that $\sum_{i=1}^l x_i = k$ and $\sum_{i=1}^l y_i = n-k$ tells us that the total weight of the coloring must be (k,n-k)$.
In the notation of the Wikipedia article, the color generating function is
$$
   f(t_1, t_2) = \frac{t_1 t_2}{(1 - t_1)(1-t_2)}
$$
in which the coefficient of $t_1^i t_2^j$ is $1$ for all $i,j\ge 1$, corresponding to how there's a single color of each weight that's positive in both coordinates.
The cycle index $Z_G(t_1, \dots, t_l)$ is a generating function for the cycle structures of the elements of the group $G$, which in this case is the $l$-element cyclic group. This is what's going to give us the most trouble. When $l$ is prime, we have $l-1$ elements which are $l$-cycles (as permutations of $\{1,2,\dots,l\}$) and one identity element, so in that case
$$
   Z_G(t_1, \dots, t_l) = \frac{t_1^l + (l-1) t_l}{l}.
$$
When $l$ is not prime, things get trickier. For example, when $l=6$, the identity still has term $t_1^6$ and rotations by $\frac\pi3$ and $\frac{5\pi}{3}$ still have term $t_6$, but rotations by $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ have term $t_3^2$ (two $3$-cycles) and rotating by $\pi$ has term $t_2^3$ (three $2$-cycles) so we get
$$
   Z_G(t_1, \dots, t_6) = \frac{t_1^6 + t_2^3 + 2t_3^2 + 2t_6}{6}.
$$
In general, if $\phi$ is the totient function, we have
$$
   Z_G(t_1, \dots, t_l) = \frac1l \sum_{d \mid l} \phi(d) t_{d}^{l/d}.
$$
Then the number of solutions is the coefficient of $t_1^k t_2^{n-k}$ in
$$
  Z_G(f(t_1, t_2), f(t_1^2, t_2^2), \dots, f(t_1^l, t_2^l)) = \frac{(t_1 t_2)^l}{l} \sum_{d \mid l} \frac{\phi(d)}{((1-t_1^d)(1-t_2^d))^{l/d}}.
$$
