TITLE: Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$
QUESTION [7 upvotes]: Find an analytic formula for the recurrent sequence $$q_{n+1}=q_n(q_n+1)+1,\;\;q_0\in\mathbb N.$$
(The question was asked on 03.05.2018 by M. Pratsovytyi, see page 109 of Volume 1 of the Lviv Scottish Book).

REPLY [9 votes]: If we denote $A_n=q_n+1/2$, then 
$$A_n=A_{n-1}^2+5/4$$
with $A_0=q_0+1/2\ge 3/2$ by $q_0\in\mathbb{N}$.
Further, 
$$\log A_n=2\log A_{n-1}+\log\left(1+\frac{5}{4A_{n-1}^2}\right),$$
namely
$$\frac{1}{2^n}\log A_n-\frac{1}{2^{n-1}}\log A_{n-1}=\frac{1}{2^n}\log\left(1+\frac{5}{4A_{n-1}^2}\right).$$
Thus
$$\log A_n=2^n\left(\log A_{0}+\sum_{k=1}^{n}\frac{1}{2^k}\log\left(1+\frac{5}{4A_{k-1}^2}\right)\right).$$
Clearly,
$$A_n> A_{n-1}^{2^1}> A_{n-2}^{2^2}>\cdots A_0^{2^n}\ge (3/2)^{2^n}.$$
Thus,
$$0<\sum_{k\ge n+1}\frac{2^n}{2^k}\log\left(1+\frac{5}{4A_{k-1}^2}\right)<\sum_{k\ge n+1}\frac{2^n}{2^k}\frac{5}{4A_{k-1}^2}<\frac{5}{4A_{n}^2}.$$
Hence note that $A_0>3/2$ we obtain that
$$1-\frac{5}{4A_n^2}<e^{-\frac{5}{4A_n^2}}\le A_n\kappa^{-2^n}<1$$
with
$$\kappa=A_0\prod_{k=1}^{\infty}\left(1+\frac{5}{4A_{k-1}^2}\right)^{\frac{1}{2^k}}$$ 
a constant depends only on $A_0$. Thus we can prove that
$$\kappa^{2^{n}}-3\kappa^{-2^{n}}<A_n<\kappa^{2^n}$$
for all $n\ge 2$ by note that $\kappa>A_0=3/2$.  For the computing of $\kappa$, it follows from above that
$$A_n^{1/2^n}<\kappa<A_n^{1/2^n}\left(1+\frac{3}{A_n^2}\right)^{1/2^n}<A_n^{1/2^n}\left(1+\frac{3}{2^nA_n^2}\right).$$