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

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!

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  • $\begingroup$ WolframAlpha can't do it, so it's probably difficult. $\endgroup$ – Arthur Sep 8 '16 at 20:00
  • $\begingroup$ Are you sure this is the recursion you want to consider and it can be solved? $\endgroup$ – Did Sep 8 '16 at 20:12
  • $\begingroup$ @Did see Marty's answer and math.stackexchange.com/questions/1918972/… $\endgroup$ – Will Jagy Sep 8 '16 at 20:28
  • $\begingroup$ @WillJagy Thanks. $\endgroup$ – Did Sep 8 '16 at 21:26
  • $\begingroup$ @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 $a_n.$ $\endgroup$ – Will Jagy Sep 8 '16 at 21:45
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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.

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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.

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