Closed form for a recursive sequence: reference request

Question

Playing around with this WolframAlpha widget, I found that the sequence satisfying $$f(n)=4f(n-1)+2f(n-1)^2$$ and $$f(1)=4$$ is given by $$f(n)=\cos(2^{n-1}\cos^{-1}(5))-1.$$ Could you tell me where I could read about the general theory leading to such a result?

Answer 1

actually, cosine is inappropriate as $f(1) = 4,$ unless you are willing to keep a bunch of complex coefficients. You want $a_{n+1} = 2 a_n^2 + 4 a_n,$ and I am going to shift so $a_0 = 4$

The recursions that come out well are $d_{n+1} = d_n^2$ and $c_{n+1} = c_n^2 - 2$

For your question, take $c_{n+1} = c_n^2 - 2$

Then take $$ a_n = \frac{c_n}{2} -1 $$ With this, it turns out $$ a_{n+1} = 2 a_n^2 + 4 a_n$$

You need to find $c_0,$ then there is a real quantity $$ c_0 = W + \frac{1}{W}, $$ after which $$c_n = W^{2^n} + \frac{1}{W^{2^n}}, $$ from $a_0 = 4$ we need $4 = \frac{c_0}{2} - 1$ or $c_0= 10,$ we may take $$ W = 5 + \sqrt{24}$$ with reciprocal $5 - \sqrt{25}$

$$c_n = \left(5 + \sqrt{24} \right)^{2^n} + \left(5 - \sqrt{24} \right)^{2^n} $$

and $$ a_n = \frac{c_n}{2} -1 $$

Answer 2

$$f_n=4f_{n-1}+2f_{n-1}^2$$ $$f_n=g_n-1 \quad \implies \quad g_n=2g_{n-1}^2-1$$ Here, we see appearing the solution (double angle formula for the cosine) $$g_n=\cos(h_n)\quad \implies \quad \cos(h_n)=\cos(2 h_{n-1})\quad \implies \quad h_n=2^{n-1}\,C$$ Therefore $$g_n=\cos(2^{n-1}\,C)\quad \implies \quad f_n=\cos(2^{n-1}\,C)-1$$ $$f_1=4 \implies \cos(C)=5\implies C=\cos ^{-1}(5)\quad \implies \quad f_n=\cos \left(2^{n-1} \cos ^{-1}(5)\right)-1$$ as given by Wolfram Alpha.

But $$\cos ^{-1}(5)=i \cosh ^{-1}(5)\quad \implies \quad \color{red}{f_n=\cosh \left(2^{n-1} \cosh ^{-1}(5)\right)}$$

$$f_7=2610701117696295981568349760414651575095962187244375364404428800$$