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
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3Hint: $\cos(2\theta) = 2\cos^2\theta - 1$ and $\cosh(2\theta) = 2\cosh^2\theta - 1$ – achille hui Jul 11 '18 at 11:54
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Solving recurrence $a_k = a_{k-1}^2 + c$ is possible (in some precise sense) only when $c=0$ or $c=-2$. Achille provided the hint for the case $c=-2$. – GEdgar Jul 11 '18 at 12:00
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If $|a_0|\le 2$, by induction $a_n=2\cos (2^n\arccos\frac{a_0}{2})$. If $|a_0|> 2$, by induction $a_n=2\cosh (2^n\operatorname{arcosh}\frac{a_0}{2})$.
J.G.
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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.