I have the following recurrence relation -an eigenvalue problem-
\begin{equation} (N+1-r) \, a_{r-1} + \frac{(m(r+1))!}{(m r)!} a_{r+1}=C_1 a_{r} \quad r\geq 1, \end{equation} with boundary conditions $a_{-1}=a_{N+1}=0$. $m$ is a positive integer $\geq 1$ and $C_1$ a constant. I know that the eigenvectors $a_r$ should be orthogonal polynomials for any $m$. For $m=1$ I get \begin{equation} (N+1-r) \, a_{r-1} + (r+1) a_{r+1}=C_1 a_{r}. \end{equation} This equation can be solved using a generating function $f(z)=\sum_{r=0}^{N} a_{r} z^{r}$. The differential equation is \begin{equation} (1-z^2)f'(z)=(C_1-Nz)f(z). \end{equation} The solution of this differential equation is $f(z)=C_2 (1+z)^{\frac{N+C_1}{2}}(1-z)^{\frac{N-C_1}{2}}$ and the eigenvectors $a_{r}$ can be written in terms of orthogonal polynomials using the Rogrigues formula for Jacobi polynomials and $f(z)$. $C_2$ is a constant to be determined by a normalization condition on the eigenvectors $a_r$, and $C_1$ are eigenvalues obtained imposing analiticity in $f(z)$.
For $m=2, 3, \dots$, it gets more complicated \begin{equation} (N+1-r) \, a_{r-1} + (2r+2)(2r+1) a_{r+1}=C_1 a_{r},\\ (N+1-r) \, a_{r-1} + (3r+3)(3r+2)(3r+1) a_{r+1}=C_1 a_{r}, \end{equation} and so on.
Could anybody give me a hint on how to get (if possible) the generating function $f(z)$ for $m\geq2$?