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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 $\ell \approx 2.7103$.

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    $\begingroup$ A very good question (+1). $\endgroup$ Sep 12, 2019 at 22:02
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    $\begingroup$ $2*\ln($ A076949 $)$ is what it converges to for $u_0 = 1$. In the OEIS, there is no closed form for this. $\endgroup$ Sep 12, 2019 at 23:05

1 Answer 1

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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, $\ell_k \sim k \log(10)$

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