Non linear recurrence relation I'm trying to solve the following recurrence equation, for $\alpha$ and $\beta$ in $\mathbb{R}$.
$$\forall n\in \mathbb{Z}, \; (\alpha+\beta n^2)\, f(n)=\frac{f(n+1)+f(n-1)}{2}$$
I know that if I had an $n$ instead of a $n^2$, I could introduce Bessel functions of the first kind using the relation $J_{n+1}(x)+J_{n-1}(x)={2n \over x} J_n(x)$, but this is not the case here...
Is there by any chance a special function defined by my recurrence relation ? 
 A: First, do the Fourier transform
$$\beta\frac{d^{2}}{d\xi^{2}}f(\xi)+(\cos(\xi)-\alpha){f(\xi)}=0$$
Then do the rescaling
$$\xi=2\eta$$
and you get the Mathieu Equation
$$\frac{d^{2}f(\eta)}{d\eta^{2}}+\Big(\frac{4}{\beta}\cos(2\eta)-\frac{4\alpha}{\beta}\Big)f(\eta)=0$$
It is solved by Mathieu functions. A good reference is e. kamke differentialgleichungen!
A: Use generating functions. Define:
$\begin{equation*}
F(z) = \sum_{n \ge 0} f(n) z^n
\end{equation*}$
Shift the recurrence to get rid of subtractions in indices,
multiply by $z^n$, sum over $n \ge 0$ and recognize some of the sums:
$\begin{equation*}
  \sum_{n \ge 0} 2 (\alpha + \beta (n + 1)^2) f(n + 1) z^n
    = \sum_{n \ge 0} f(n + 2) z^n
         + \sum_{n \ge 0} f(n) z^n \\
\end{equation*}$
Use the facts:
$\begin{align*}
   \sum_{n \ge 0} f(n + r) z^n
     &= \frac{F(z) - f(0) - f(1) z 
                   - \cdots - f(r - 1) z^{r - 1}}
              {z^r} \\
   \sum_{n \ge 0} n^r f(n) z^n
     &= \left( z \frac{d}{d z} \right)^r F(z)
\end{align*}$
and you get a differential equation in $F(z)$. Solve that one (good luck!), it's Maclaurin series gives your solution as it's coefficients.
