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If $a_{n+1}=a_n^2+1,$ with initial $a_1=\frac{1}{2}$. How to solve this sequence problem, i.e., how to represent $a_n$ in closed form?

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    @J.D.Than$k$s $f$or the correction. I immediately regretted my vote to close, but the so$f$tware does not allow me to retract it. Fortunately there is no harm done.2012-05-16

3 Answers 3

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EDIT: The answer below assumes $a_1$ is an integer, and as such is not directly applicable to your question, but I am leaving it here in case people can make use of the results mentioned in the paper below.

Aho and Sloan proved that for sequences like yours, there is a constant $k$ such that

$a_n = \lfloor k^{2^n} \rfloor$

for sufficiently large $n$. $k$ can be defined as a limit of a sequence using $a_n$ itself. If you include $k$ as one of your closed form constants, you are done!

See their paper for details: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf

Of course, for your special case, one might still be able to find a different "closed form" which might be more appealing to you.

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    @Didier: I added a disclaimer! It might need some work to get those results to work for this problem (if at all they work). But I have left the answer, in case someone wants to try.2012-05-16
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The recurrence $a_{n+1} = a_n^2+c$ has a (known) closed form if and only if $c=0$ or $c=-2$. See this answer for more explanation.

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Perhaps let $a_n=\tan \theta_n$. Then $\tan \theta_{n+1}=\sec^2 \theta_n$. So, $\tan \theta_2=\sec^2 \theta_1$, $\tan \theta_3=1+\sec^4 \theta_1$, etc. I am not sure if this procedure will produce a closed form.