There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.
However super-half-exponentials (functions whose composition grows faster than exponential) have lot of candidates.
Eg: $2^n$ itself.
A little thought gives $f(0,a,n)=2^{n^a}$, $f(1,a,n)=2^{2^{(\log n)^a}}$, $f(2,a,n)=2^{2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)=2^{2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(0,1)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.
The functions grow slower as $k$ increases. Define $g_{upper}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.
- Is $g_{upper}(a,n)$ almost the half-exponential function? That is, does $g_{upper}(a,g_{upper}(a,n))=2^{O(n^{1+\epsilon})}$ hold at every $a\in(0,1)$ and at every $\epsilon>0$?
I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s decrease as $k$ increases. It is possible $g_{upper}(a,n)$ is the half-exponential function.
It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{lower}(a,n)$ is defined in Natural candidates for sub-half-exponential which limit to half-exponential function from below.
That is $g_{upper}(a,g_{upper}(a,n))=2^{O(n)}$ holds which means $g_{upper}(a,n)$ is the half-exponential function.
- Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k+1,a',n)$ where $a'\in(a,1)$) with $$f(k+1,a,n)\ll h(k,n)\ll f(k,a,n)$$ $$h(k,h(k,n))=2^{O(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?
