Closed form for solution of $t_{n+1}=t_n(t_n-2)$

Question

As in the title I am interested in finding closed form for sequence satysfing $$t_{n+1}=t_n(t_n-2)$$ with $t_1=4$. I have tried many guesses, because I don't know if there is a metod to solve that, but without succes.

Answer 1

By setting $t_n=a_n+1$ we have: $$ a_{n+1} = a_n^2-2 \tag{1}$$ then by setting $a_n=2b_n$ we have: $$ b_{n+1} = 2 b_n^2-1 \tag{2} $$ It is not difficult to recognize the duplication formula for the (hyperbolic) cosine.

In particular, assuming $b_0=\cosh t_0$, $b_n = \cosh(2^n t_0)$ follows by induction.

Since in our case $b_1=\frac{3}{2}$ gives $b_0=\frac{\sqrt{5}}{2}$, we have $t_0=\log\left(\frac{1+\sqrt{5}}{2}\right)=\log\phi$, hence:

$$ t_n = 1+2\cosh(2^n \log\phi) = 1+\left(\phi^{2^n}+\phi^{-2^n}\right)=1+L_{2^n}\tag{3} $$

where $L_m$ is the $m$-th Lucas number.

Answer 2

Hint: prove by induction that: $$t_n=\alpha^{2^n}+\alpha^{-2^n}+1 $$

with $\alpha$ is one of the roots of $x+\frac{1}{x}=3$