I am having trouble with the following recurrence relation:
$$c_{n+1} - c_{n-1} = 2\alpha \sin \frac{(2n-1)\pi}{N} c_n, \quad\forall n \in \mathbb{Z},$$
where $N$ is odd and the initial condition is $c_0 = c_1 = 1$.
Solving explicitly for $c_2$, $c_3$, ..., it is obvious that $c_k$ is a polynomial in $\alpha$ with a maximum degree of $k-1$. A closer analysis reveals that its degree reaches its maximum up to $k = \lfloor N/2 \rfloor+1$, after which the highest order terms start cancelling out one by one, recovering previous values in reverse order and leaving $c_N = c_{N+1} = 1$ again. Hence I know is that the solution is periodic with a period of $N$ and symmetric with respect to $n \mapsto 1-n$. Not much more, however. Rewriting the problem in the discrete Fourier image provides no help as—quite interestingly—the form of the equation stays exactly the same (only the value of $\alpha$ becomes reciprocal).
Can anyone help shed some light on finding a closed form solution of this kind of problem? It seems to be related quite tightly to the analysis of discrete Fourier transform eigensystem but that is beyond my current knowledge.