Find a closed form for the sequence $a_{n+1}= \alpha + \beta a_n^\gamma$ Let $\left(a_k\right)_{k\geq 0}$ s.t. :
$$a_{n+1}= \alpha + \beta a_n^\gamma.$$
Is there any closed form ? If not, is there a closed form when $\gamma=-1$ ? ($a_{n+1}=\alpha + \frac{\beta}{a_n}$)
 A: For general $\gamma$ the answer is no. Otherwise, setting $\gamma=2$, one would be able to write a closed form solution for the logistic map.
As for $\gamma=-1$, there is a closed form, since one can represent this map by a fractional-linear transformation:
$$a_{n+1}=\frac{\alpha\cdot a_n+\beta}{1\cdot a_n+0}.$$
Hence we can write $a_n=\frac{x_n}{y_n}$, where
$$\left(\begin{array}{c}
x_n \\ y_n\end{array}\right)=M^n\left(\begin{array}{c}
x_0 \\ y_0\end{array}\right),\qquad M=\left(\begin{array}{cc}\alpha & \beta \\ 1 & 0\end{array}\right).$$
It now suffices to diagonalize $M$ to be able to write the general solution. Namely, denoting $\displaystyle \lambda_{\pm}=\frac{\alpha\pm\sqrt{\alpha^2+4\beta}}{2}$, we have $M=P\left(\begin{array}{cc}\lambda_- & 0 \\ 0 & 
\lambda_+\end{array}\right)P^{-1}$, with
$\displaystyle P=\left(\begin{array}{cc}\lambda_- & \lambda_+ \\ 1 & 
1\end{array}\right)$, and therefore
$$M^n=P\left(\begin{array}{cc}{\lambda_-}^{\!\!\! n} & 0 \\ 0 & 
\lambda_+^n\end{array}\right)P^{-1}=\frac{1}{\lambda_--\lambda_+}
\left(\begin{array}{cc}{\lambda_-}^{\!\!\! n+1}-\lambda_+^{n+1} & -\lambda_-\lambda_+\left({\lambda_-}^{\!\!\! n}-\lambda_+^{n}\right) \\ {\lambda_-}^{\!\!\! n}-\lambda_+^{n} & 
-\lambda_-\lambda_+\left({\lambda_-}^{\!\!\! n-1}-\lambda_+^{n-1}\right)\end{array}\right).$$
We thereby obtain
$$a_n=\frac{\left({\lambda_-}^{\!\!\! n+1}-\lambda_+^{n+1}\right)a_0
-\lambda_-\lambda_+\left({\lambda_-}^{\!\!\! n}-\lambda_+^{n}\right)}{\left({\lambda_-}^{\!\!\! n}-\lambda_+^{n}\right)a_0
-\lambda_-\lambda_+\left({\lambda_-}^{\!\!\! n-1}-\lambda_+^{n-1}\right)}.$$
