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There is no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ almost the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n^{1-\epsilon})}$ 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 increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above.

That is $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ holds which means $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ almost the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n^{1-\epsilon})}$ 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 increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above.

That is $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ holds which means $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

There is no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ hold at every $a\in(0,1)$?

I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above

There is no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ almost the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n^{1-\epsilon})}$ 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 increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above.

That is $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ holds which means $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

There is no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ hold at every $a\in(0,1)$?

I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

There is no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ hold at every $a\in(0,1)$?

I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above