Simplifying a ‘fractal-like’ expression with tetration

big numbersclosed-formhyperoperationpower-towerstetration

Let $f_2(n)=2^n n$ and let $f_3$ be defined recursively as
$$
f_3(n)=\underbrace{f_2\cdots f_2}_{n\text{ times}}(n)=f_2^n(n).
$$
This will lead to tetration, but is it possible to write $f_3$ in a closed formula, using the notation ${^{n}a}$ for tetration (or even Knuth's up-arrow notation)?

I tried to write a simple code to see an emerging pattern, but simplifications seem somehow tricky. Also, ${^{n}2}$ seems to be relatively simple to recognise in the expansion of $f_3$, but the remaining terms are somewhat complicated and it almost feels like the expression behaves like a fractal. Any ideas?

Here are the first four iterations, might help to picture the pattern.
\begin{align*}
f_2(n)&=2^n n\\
f_2^2(n)&=2^{(2^n+1)n}n=2^{2^nn}2^nn\\
f_2^3(n)&=2^{(2^{(2^{n}+1)n}+2^n+1)n}n=2^{2^{2^nn}2^nn}2^{2^nn}2^nn\\
f_2^4(n)&=2^{(2^{(2^{(2^n+1)n}+2^n+1)n}+2^{(2^n+1)n}+2^n+1)n}n
\end{align*}
Just to give some context, these functions were defined and discussed in a recent Numberphile video about $\text{TREE}(n)$ and Graham's number (see around minute $10$).

Best Answer

The simple answer is no, because there's a mix of operations being iterated: either multiplication and exponentiation, or if you bring the multiplication up a level, addition and exponentiation.

As far as approximations go though, you can see here for some tight bounds:

$$n\operatorname{Tet}(2^n,1,n)\le f_3(n)\le\operatorname{Tet}(2\sqrt[n]n,n,n)$$

where $\operatorname{Tet}(a,b,c)=\underbrace{a~\widehat~~a~\widehat~~\dots~\widehat~~a~\widehat~~}_cb$ is $c$ powers of $a$ with $b$ on top.

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