Recurrence $a_n = \sum_{k=1}^{n-1}a^2_{k}, a_1=1$

Question

This seems like a really straightforward recurrence.

I wrote out the first few terms: $1,1,2,6,42,1806$... It seems to grow faster than $n!$ but slower than $n^n$. Any suggestions about the closed form or the rate of growth are very welcome.

Answer 1

$$ a_{n+1} = a_n (1+a_n) $$ ..................

So, with or without closed form, there is some positive real $\theta$ such that $$ a_n \approx \theta^{\left( 2^n \right)} $$

and $\theta $ can be approximated by $a_n^{1/2^n},$ while $\log \theta$ can be approximated by $\frac{\log a_n}{2^n}. $

EEddIItt: Convergence for $\theta$ should be rapid. I get that $\frac{\log a_n}{2^n} $ increases with $n,$ however $$ \frac{\log a_{n+1}}{2^{n+1}} - \frac{\log a_n}{2^n} < \frac{\log \left(1 + \frac{1}{2 a_n} \right)}{2^n} $$

I numbered my version with $$ a_1 = 6, \; a_2 = 42, \; a_3 = 1806. $$ With this numbering I get $\theta \approx 2.553317.$ Anyway, just look at the logarithms, this grows much faster than $n^n$