Let $u_0 \in \mathbb{N}^{*}$ and $(u_n)$ such that $u_{n+1}=1+u_n^2$.
I've shown that the sequence $\displaystyle \left(\frac{\ln\left(u_n\right)}{2^n}\right)$ converges and I wonder if it possible to find its limit $\ell$
For example, if $u_0=1$ then $\ell \approx 0.40735$, if $u_0=15$ then $\ell \approx 2.7103$.
This is not an answer.
With the $2^n$ in denominator, the limit is obtained quite fast.
Let $u_0=10^k$ and computing we get $$\left( \begin{array}{cc} k & \ell_k \\ 0 & 0.407354523 \\ 1 & 2.307584766 \\ 2 & 4.605220186 \\ 3 & 6.907755779 \\ 4 & 9.210340377 \\ 5 & 11.51292547 \\ 6 & 13.81551056 \\ 7 & 16.11809565 \\ 8 & 18.42068074 \\ 9 & 20.72326584 \end{array} \right)$$ and, as you can see, $\ell_k \sim k \log(10)$
