Rim hook decomposition and volume of moduli spaces I did some computer experiments, counting the number of rim-hook decompositions (aka border-strip decompositions) of rectangles of shape $2n \times n$, where each strip has size $n$.
Here are 12 of the 1379 such decompositions for $n=4$:

Let $a_n$ be the number of such decompositions.
I conjecture that $a_n$ is given by A115047 in OEIS,
that is, $a(0)=1$ and
$$
a(n) = \sum_{i=1}^n
   \binom{n - 1}{i - 1} \binom{n + 3}{i + 1} a(i - 1) a(n - i) \frac{ i(n - i + 1)}{2 (n + 2)},
$$
or $1, 5, 61, 1379, 49946, 2648967,...$. 
Edit(2018-02-22): My student have checked that the sequences agree up to $n=12$, which is quite compelling.
The interesting part is that there is no such reference in OEIS. This entry regards Weil-Petersson volumes of muduli spaces of an $n$-punctured Riemann sphere, which is quite far from my field.
Q: Can we prove the recursion above?
I think this (conjectured) connection is interesting, giving a combinatorial interpretation of moduli space volumes. 
Perhaps this can be extended to other genus?
EDIT: Using the strategy in the formula given in the answer below,
this conjecture is now proved in this preprint.
Edit II Paper is now published in the Journal of Integer sequences.
The answer to the original question is therefore, YES.
 A: Abacus is quite useful to count objects related to hooks. For partition $((2n)^n)$, the $n$-abacus is $\{2n, \dots, 3n-1\}$ on $n$ runners. Thus, the number of labelled rim hook decompositions (labelled by the order of removal) equals to the number of permutations of $x_1, \dots, x_n, y_1, \dots, y_n$ such that $x_i$ appears before $y_i$ for every $i$.
In order to count unlabelled rim hook decompositions, we need to find out when the order of two consecutive rim hook removal can be swapped. Then, we can add some constraint that allows one type of removal, but forbids the other type of removal. All labelled rim hook decompositions with the new constraint are in bijection with unlabelled rim hook decompositions.
If we translate the new constraint we get to permutations of $x_i$'s and $y_i$'s, that means: the permutation does not have consecutive $x_i$ and $y_j$ such that $i \gt j$.
It is then straightforward to use inclusion-exclusion to count the number of permutations satisfying both requirements. I get a complicated formula for $a_n$, and the formula matches A115047 for $n$ up to $60$. I believe that it should not be too hard to prove that the formula for $a_n$ satisfies the recursion.

Edit (Mar. 3, 2018)
The $n$-abacus of a partition $\lambda = (\lambda_1, \dots, \lambda_m)$ consists of $n$ runners and $m$ beeds located at $\{\lambda_i + m - i\}$, and the map from beeds to runners is given by $\mathbb{N} \to \mathbb{N} / n \mathbb{N}$. Removing an $n$-hook corresponds to the move of a beed along its runner to a smaller adjacent unoccupied location. A labelled rim hook decomposition is just a sequence of such moves so that no further moves can be made.
For the partition $((2n)^n)$, let $x_i$ be the move of the beed at $2 n + i - 1$ to $n + i - 1$, and let $y_i$ be the move of the beed at $n + i - 1$ to $i - 1$. Two hook removal at $b$ and $b'$ can be swapped if and only if $|b - b'| \gt n$. For this particular partition, that means $x_i y_j$ or $y_j x_i$ such that $i \gt j$. Therefore, we can reformulate the problem of finding $a_n$ to a problem of counting permutations of $x_i$'s and $y_i$'s. For general partitions, the same method works, and we just need to consider possibly larger "alphabet" and longer "forbidden words".
Below is one method of counting such permutations. There must be a much better way to count them.
Suppose that we know some consecutive occurrences of $x_i y_j$'s. Consider the bipartite graph with vertices $x_i$'s and $y_i$'s and edges $x_i y_i$ for all $i$ and $x_i y_j$ for all known consecutive $x_i y_j$ with $i \gt j$. This bipartite graph is a disjoint union of paths, otherwise it would not be legitimate. Let $p$ be the partition such that the parts of $2p$ are the sizes of the connected components of the bipartite graph mentioned above. Therefore, by inclusion-exclusion, we have the formula
$$ a_n = \sum_{p \vdash n} (-1)^{|p - 1|} \frac{1}{|m|!} {|m| \choose m} {|p| \choose p} {|p + 1| \choose p + 1},$$
where $p$ runs over partitions of $n$, $m := (m_1, \dots, m_n)$ with $m_i$ being the multiplicity of $i$ in $p$, $p + c$ is the addition of $c$ to each part of $p$, $|p|$ is the sum of parts in $p$ and ${|p| \choose p}$ is the multinomial coefficient.
