TITLE: What combinatorial quantity the tetration of two natural numbers represents?
QUESTION [26 upvotes]: 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.

REPLY [4 votes]: $2 \uparrow \uparrow (n-1)$ 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
$$V_\alpha = \bigcup_{\beta < \alpha} \mathcal{P}(V_\beta)$$