What combinatorial quantity the tetration of two natural numbers represents?

Tetration is a generalization of exponentiation in arithmetic and a part of a series of other generalized notions, Hyperoperators. Consider $m\uparrow n$ denotes the tetration of $m$ and $n$. i.e. $$\underbrace{m^{m^{m^{.^{.^{.^{m}}}}}}}_{n-times}$$

Note that one can find a combinatorial description of each one of operators sum, multiplication and exponentiation as follows:

• $m+n$ is the size of disjoint union of two sets with $m$ and $n$ elements.

• $m.n$ is the size of Cartesian product of two sets with $m$ and $n$ elements.

• $m^n$ is the size of set of all functions from a set with $n$ elements to a set with $m$ elements.

• $m\uparrow n$ is the size of ... (?)

Question: Is it possible to introduce a combinatorial set (defined by $m$, $n$) which its size is $m\uparrow n$ as well as the case of $m+n$, $m.n$, $m^n$? What about other Hyperoperators like pentation and hexation? The simple and most natural expressions are more interesting.

This is probably not what you want. but we can create the notion of the $n$'th dual of a set (I'm borrowing this notation from linear algebra).

Given sets $X$ and $Y$ consider the dual of $X$ with respect to $Y$ to be the the functions from $X$ to $Y$.

Define recursively the $n$'th dual of a set $X$ to be the dual of $X$ with respect to $X$ for $n=1$ and the dual of the $n-1$'th dual of $X$ with respect to $X$ for $n>1$. Then $m\uparrow n$ is the order of the $n$'th dual of a set of $m$ elements.

$2 \uparrow \uparrow (n-1)$ is the number of sets of rank $n$. Intuitively, the rank of a set is the depth of the nesting braces needed to write it. Recursively, the rank of the empty set is zero, and the rank of a nonempty set is the one plus the maximum rank of its elements. It is also the depth of the stack needed by a pushdown automaton to recognize the braces are balanced. In short, $2 \uparrow \uparrow (n-1) = |V_n|$ where

$$V_\alpha = \bigcup_{\beta < \alpha} \mathcal{P}(V_\beta)$$