calculus - Find $a_{n+1}=frac{a_n^2+1}{2}$ in terms of $n$.

I was trying to prove that for all $n\in \Bbb N$ there are integer numbers $\{a_1,a_2,\ldots,a_n,b_n\}$ s.t. $a_1^2+a_2^2+\dots+a_n^2=b_n^2$.
I founded that if $\{a_1,a_2,\ldots,a_n,b_n\}$ have the property then $\{a_1,a_2,\ldots,a_n,a_{n+1}=\dfrac{b_n^2-1}{2},b_{n+1}=\dfrac{b_n^2+1}{2}\}$ are the $n+1$ numbers which have the property too. I start with $a_1=b_1=3$ and get the sequence $3,4,12,84,\ldots$ and now try found $a_n$ in terms of $n$. I also prove that solving $b_{n+1}=\dfrac{b_n^2+1}{2},b_1=3$ is equivalent to solving $c_{n+1}=c_n^2+c_n+1 , c_1=1$. Thanks for reading the text and any helps.
NOTE: It's eaasy to check that for $b_1=3$ all $a_i$ and $b_i$ will be integer.

Answer

Let us tackle $c_{n+1} = c_n^2 + c_n + 1$

Set $x_n = c_n + \frac{1}{2}$.

We get

$$ x_{n+1} = x_n^2 + \frac{5}{4}$$

This paper: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf (see page 434) shows that $x_n$ is of the form: (nearest integer to $k^{2^n} + \frac{1}{2}$) - $\frac{1}{2}$ for all large enough $n$, for some constant $k$ (defined in terms of $x_n$ itself).

and thus $$c_n = \left\lceil k^{2^n} + \frac{1}{2} \right\rceil$$

It is unlikely you will get a "neater" closed form.