Cfrgtkky

I need to find out how many different colorings you can make with 2 colors in a dihedral group $D_n$ with $n$ prime and $m$ black and $p-m$ white beads. So first I compute the cycle index:
The cycle index of a dihedral group with $n$ prime (odd) is equal to:
$$Z(D_n) = frac{1}{2}(frac{1}{n}a_1^n + frac{(n-1)}{n}a_n + a_1a_2^frac{n-1}{2})$$
Now I fill in:
$$a_1 = (b+w), a_2 = (b^2 + w^2), a_n = (b^n + w^n)$$
After that, I find the number before the $b^mw^{p-m}$ and that is the amount of different colorings with $m$ black and $p-m$ white beads. But is there a general formule to find that number?

Cycle index.

$$Z(D_p) = frac{1}{2p}
left(a_1^{p} + (p-1) a_p + p a_1 a_2^{(p-1)/2}right)$$

We are interested in

$$[B^m W^{p-m}] Z(D_p; B+W).$$

This has three components.

First component.

$$[B^m W^{p-m}] frac{1}{2p} (B+W)^p
= frac{1}{2p} {pchoose m}.$$

Second component.

$$[B^m W^{p-m}] frac{p-1}{2p} (B^p+W^p).$$

This is using an Iverson bracket:

$$frac{p-1}{2p} [[m=0 lor m=p]].$$

Third component.

$$[B^m W^{p-m}] frac{1}{2} (B+W) (B^2+W^2)^{(p-1)/2}.$$

Now with $p$ prime we cannot have both $m$ and $p-m$ even, or both odd,
so one is odd and the other one even. Supposing that $m$ is odd we get

$$[B^{m-1} W^{p-m}] frac{1}{2} (B^2+W^2)^{(p-1)/2}
\ = [B^{(m-1)/2} W^{(p-m)/2}] frac{1}{2} (B+W)^{(p-1)/2}
= frac{1}{2} {(p-1)/2 choose (m-1)/2}.$$

Alternatively, if $p-m$ is odd we get

$$[B^{m} W^{p-m-1}] frac{1}{2} (B^2+W^2)^{(p-1)/2}
\ = [B^{m/2} W^{(p-m-1)/2}] frac{1}{2} (B+W)^{(p-1)/2}
= frac{1}{2} {(p-1)/2 choose m/2}.$$

Closed form.

$$bbox[5px,border:2px solid #00A000]{
frac{1}{2p} {pchoose m}
+ frac{p-1}{2p} [[m=0 lor m=p]]
+ frac{1}{2} {(p-1)/2 choose (m-[[m ;text{odd}]])/2}.}$$

Sanity check.

With a monochrome coloring we should get one as the answer, and
we find for $m=0$ ($B^0 W^p = W^p$)

$$frac{1}{2p} {pchoose 0} + frac{p-1}{2p}
+ frac{1}{2} {(p-1)/2 choose 0}
= frac{p}{2p} + frac{1}{2} = 1.$$

Similarly we get for $m=p$ ($B^p W^0 = B^p$)

$$frac{1}{2p} {pchoose p} + frac{p-1}{2p}
+ frac{1}{2} {(p-1)/2 choose (p-1)/2}
= frac{p}{2p} + frac{1}{2} = 1.$$

The sanity check goes through. Another sanity check is $m=1$ or
$m=p-1$ which should give one coloring as well. We find

$$frac{1}{2p} {pchoose 1}
+ frac{1}{2} {(p-1)/2choose 0} = 1$$

and

$$frac{1}{2p} {pchoose p-1}
+ frac{1}{2} {(p-1)/2choose (p-1)/2} = 1.$$