Consider the following problem:
$$\tau_{0}=0\qquad\tau_{k+1}=e^{-1+\tau_k}\qquad f(k)= 1 - \tau_k$$
Does $f(k)$ have any asymptotic estimate when $k$ get large, say $k=2^{64}$?
I have run an computer program to compute $f(k)$ for $\log_2(k)\in \{10, 11, \cdots,29\}$, the results shows that $-\log_2(f(k))\sim \log_2(k)-1$. See the figure of $-\log_2(1-\tau_k)$ for $\log_2(k)\in \{10, 11, \cdots,29\}$:
There is some weird behavior for $-\log_2(1-\tau_k)$ when $\log_2(k)\geq 28$. It stable at the same value for $\log_2(k)\geq 28$. I guess it was because the computer program was written in C++ and there exist some precision problem.
Could anybody help to provide an asymptotic estimate for $f(k)$ when $k$ get large, say $k \geq 2^{64}$? Any kind of reference would also be appreciated.
PS: I found a paper named On the height of trees in which one can find a precise estimate on the ratio $ \tau_k $ which is as follows, $$ \tau_k \approx_{k \rightarrow \infty} 1 - \frac{2}{k} + \frac{2}{3}\frac{\log k}{k^2} + \frac{c}{k^2} + \cdots$$ where $c$ is a certain constant.
It suggests that $ \lim_{k\rightarrow \infty} k f(k) = 2 $. Unfortunately, the authors did not provide the proof. It seems that this formula was got by study the regular iteration of real and complex functions.