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Jun 29, 2020 at 19:12 comment added VS. I think your argument works only for finite $k$.
Jun 29, 2020 at 18:20 comment added VS. It is not clear whether 'suggests' is a formal proof since induction works for every finite $k<\infty$. What we have is a limit and I think we need to describe how fast we approach the goal to be formal. Is what you have a formal proof or just an intuition that is intended to be extrapolated somehow?
Jun 29, 2020 at 18:17 comment added Gerald Edgar My argument suggests that the limit when $k \to \infty$ is still far short of half-exponential.
Jun 29, 2020 at 18:12 comment added VS. Ok so at $k\rightarrow\infty$ we will get half-exponential?
Jun 29, 2020 at 14:15 comment added Gerald Edgar We use $a>1$ so $an>n$. So $an < n^a < \exp((\log n)^a) < \exp(\exp((\log\log n)^a))$ and so on (for large $n$).
Jun 29, 2020 at 13:37 history edited Gerald Edgar CC BY-SA 4.0
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Jun 29, 2020 at 13:30 comment added VS. So $f(k,a,n)$ is not a sub half exponential function that approaches the half exponentials in the limit from below in limit $k\rightarrow\infty$? Instead of $2$ anything larger than $1$ is fine. I meant $n$ grows and so being complex is not considered.
Jun 29, 2020 at 13:24 history answered Gerald Edgar CC BY-SA 4.0