Solving $a_k=a_{k-1}^2 -2$

Question

How do I solve this recurrence relation for a given $a_0$? I only know how to do linear ones by substituting $a_k=\lambda^k$ and solving for $\lambda$ but this doesn't work here

Answer 1

This paper contains the very general answer for your question. Because your $g_n$ (following the notation therein) is always negative, the closed form is simply $[k^{2^n}]$ for some $k$ and large enough $n \geq n_0$. Also, $k$ will depend on $a_0$, and if $a_0$ is (positive) integer, determining $k$ becomes quite easy.

Answer 2

If $|a_0|\le 2$, by induction $a_n=2\cos (2^n\arccos\frac{a_0}{2})$. If $|a_0|> 2$, by induction $a_n=2\cosh (2^n\operatorname{arcosh}\frac{a_0}{2})$.