TITLE: How can I find the closed form of this recursive relationship: $a_{n}=(a_{n-1})^2+a_{n-1},a_{0}=1$
QUESTION [5 upvotes]: This comes up in OEIS as A007018. However the recursive form is useless to me, I need the closed form. I've been trying for several hours and I simply come up empty. Any advice?
Thanks.

REPLY [2 votes]: There are just two (diagonalized monic) versions that give nice closed forms:
$$ x_n = x_{n-1}^2 $$
gives $$ x_n = x_0^{\left( 2^n \right) }.  $$
The other, used by Lucas, is
$$ x_n = x_{n-1}^2 - 2; \; \; \; \;  x_0 > 2. $$ 
This time we find $A > 1,$ with $AB = 1$ and $A + B = x_0.$
Then
$$ x_n = A^{\left( 2^n \right) } + B^{\left( 2^n \right) }.  $$
That is all the nice ones. For the others, taking logarithm of both sides shows that there is a limit which is the number $c$ from OEIS, but we can only estimate $c$ by calculating many terms of the sequence itself. 
With your sequence, taking $$ a_n = b_n - \frac{1}{2} $$
gives
$$  b_n = b_{n-1}^2 +  \frac{1}{4} $$ 
so you are out of luck as far as closed form answers. With this much, not difficult to prove that
$$  \frac{\log b_n}{2^n}  $$ has a limit as $n \rightarrow \infty,$ call it $w,$ then the number $c = e^w.$