Recreational checkers problem

Interpreting white checkers as 0s and black checkers as 1s, the checker that goes between 2 checkers is their sum mod 2. (Also known as the XOR of their values.) We can interpret the state of the ring as a vector, and the function that takes us from one ring to the next is a linear function (mod 2). Take the basis vectors to be those with a 1 in one place on the ring, and 0s everywhere else. If we can find out the fate of these basis vectors, we can find out the fate of the entire ring by summing the fates of the basis vectors corresponding to the initial configuration of the ring. Since the ring can be rotated without changing the rules, we really only need to find the fate of just one of the basis vectors, all the others will be the same.

Let $omega=e^{2pi i/N}$. We'll interpret our vectors as polynomials in $omega$. (The fact that the ring loops around is taken care of by the fact that $omega^N = 1$.) Going from one state to the next corresponds to multiplying by $(1+omega)$. Our basis vector is just equal $1timesomega^0=1$. Updating $n$ times corresponds to multiplying by $(1+omega)^n$. After $N$ steps, we will have $(1+omega)^N times 1$. By the binomial theorem, this equals:

$$ sum_{k=0}^N binom{N}{k}omega^k = 1 + omega^N + sum_{k=1}^{N-1} frac{N!}{k!(N-k)!}omega^k $$

$$ = 2 + sum_{k=1}^{N-1} frac{N!}{k!(N-k)!}omega^k $$

Now suppose that $N$ is a power of 2, that is $N=2^n$. Then we have:

$$ 2 + sum_{k=1}^{2^n-1} frac{2^n}{2^n-k}frac{(2^n-1)!}{k!(2^n-k-1)!}omega^k = 2 + sum_{k=1}^{2^n-1} frac{2^n}{2^n-k}binom{2^n-1}{k-1}omega^k $$

We can then see that in the sum, the coefficients of the $omega^k$ will all be even. So, mod 2, they will all be equal to 0. Similarly, the coefficient of $omega^0$ is 2, which is also equal to 0 mod 2. So the result will be all 0s (i.e. all white checkers).

If $N$ is not a power of 2, then there will be some odd binomial coefficients in the sum, so the vectors will not go to 0 after $N$ steps at least, and maybe not ever.