$a_{n+1}=3a_n^2+2$,$a_1=1$
I want to do $log$ to kill the square but I don't how
I can let the coefficient 3 be 1
$\frac{a_{n+1}}{3^{2^{n+1}-1}}=(\frac{a_n}{3^{2^n-1}})^2+\frac{2}{3^{2^n-1}}$
But it seems more complicated
I don't know how to do it
Thanks!
On
General method of solving such recurrence relations was designed by S.Rabinovich, G.Berkolaiko and S.Havlin. This method gives solution in the following form: $$ a_{n}=\langle e|T^n|\gamma \rangle $$ where $\langle e|=[\delta_{j1}]^\infty_{j=0}-$row-vector, $|\gamma\rangle=\{1\}^\infty_{j=0}-$column-vector, $T-$matrix, which elements defined as $T_{jk}=\binom{j}{k/2}3^{\frac{k}{2}}2^{j-\frac{k}{2}}$.
However, I should admit that derivation of simple expression for $T^n$ is quite competitive task.
A very partial start, not in any way complete.
To get rid of the $3$, let $a_n =c b_n $. Then $a_{n+1}=3a_n^2+2 $ becomes $cb_{n+1} =3(cb_n)^2+2 =3c^2b_n^2+2 $ or $b_{n+1} =3cb_n^2+2/c $.
Setting $c = \frac13$, this becomes $b_{n+1} =b_n^2+6 $.
At this point, we get into nonlinear recurrences, which generally grow exponentially.
So I'll leave it at this.