# Finding general term for sequence

$a_{n+1}=3a_n^2+2$,$a_1=1$

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!

• WolframAlpha can't do it, so it's probably difficult. – Arthur Sep 8 '16 at 20:00
• Are you sure this is the recursion you want to consider and it can be solved? – Did Sep 8 '16 at 20:12
• @Did see Marty's answer and math.stackexchange.com/questions/1918972/… – Will Jagy Sep 8 '16 at 20:28
• @WillJagy Thanks. – Did Sep 8 '16 at 21:26
• @Did I guess you changed your comment after the system pasted it to my "unread inbox." I suppose i was not clear; I think the two students are working on the same problem. To the best of my knowledge, only two such problems have closed form solutions, $x_{n+1} = x_n^2$ and $y_{n+1} = y_n^2 - 2.$ For any other, all that can be done is to estimate $C$ in $a_n = C^{2^n},$ but it is not even possible to get $C$ particularly accurate with only a handful of sequence terms, and a tiny error in $C$ makes an enormous difference in $a_n.$ – Will Jagy Sep 8 '16 at 21:45

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.

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.