Closed form of the sequence $a_{n+1}=a_n^2+1$ 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?
 A: 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.
A: 
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.
A: 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.
