I've found the following claim in a slightly different form here (page 4, bottom of the left column)
Starting from $x_0\le n/3$, the recurrence equation $$3\le x_{k+1}\le\left\lfloor3\frac{x_k^2(1+\delta)}n +3\right\rfloor$$ converges to $x_n=3$ after $n=\cal O(\log\log n)$ steps.
I've tried to prove this statement but I found on Wolfram that the recurrence equation $$ x_{k+1}=ax_k^2+bx_k+c $$ is not solvable in a closed form in general. (Even $x_{k+1}=x_k^2+c$ is not.)
However it is not difficult to convince myself of the soundness of the stated claim: indeed, $x_{k+1}=ax_k^2$ has a solution of the form $x_k=K(ax_0)^{2^k}$. On Wolfram again, the equation $x_{k+1}=x_k^2+1$ starting from $x_0=1$ has as solution $x_k=c^{2^k}$ and $x_{k+1}=x_k^2-x_k+1$ starting from $x_0=1$ has as solution $x_k=d^{2^k}+1/2$.
So it seems realistic to conjecture that when a recurrence equation of the form $x_{k+1}=ax_k^2+bx_k+c$ decreases strictly, it converges to $x_n=1$ in $n=\cal O(\log\log x_0)$ steps.
Is there a result like this?