# Super Catalan Numbers

In 1874, a French and Belgian mathematician Eugène Catalan introduced the numbers

$S(m,n) \equiv \dfrac{(2m)!(2n)!}{m!n!(m+n)!},$

now better known as super Catalan numbers, and stated that the numbers are integers. The entire text of his note in French, an item in a column called Questions which appeared in Nouvelles Annales de Mathématiques Volume 13, is as follows:

$a, b$ étant deux nombres entiers quelconques, la fraction

$\dfrac{(a+1)(a+2)\ldots 2a (b+1)(b+2)\ldots 2b}{1 \cdot 2 \cdot \ldots \cdot (a+b)}$

est égale à un nombre entier.

Unfortunately, there is also another sequence of numbers which goes by the name of super Catalan numbers too, called the Schröder-Hipparchus numbers. It makes the naming a bit ambiguous.

Nevertheless, $S(m,n)$ can be viewed as a generalization of the famous Catalan numbers

\begin{aligned} \mathcal{C}_n = \frac{1}{n+1}\binom{2n}{n},\end{aligned}

because $S(1,n) = 2\mathcal{C}_n.$ You can see the Catalan sequence here in Online Encyclopedia of Integer Sequences (OEIS).

You can prove that $S(m,n)$ is always an integer for any non-negative integers $m$ and $n$ by observing that $S(m,n)$ satisfies

$4S(m,n) = S(m+1,n) + S(m,n+1),$

and that $S(m,0)$ is the central binomial coefficient $\displaystyle\binom{2m}{m} \in \mathbb{Z}$, where we can proceed by induction on $n$. Except for $S(0,0) = 1$, $S(m,n)$ is always even so sometimes $\dfrac{1}{2}S(m,n)$ is considered.

Unlike the Catalan numbers which so far have more than 200 combinatorial interpretations (Richard Stanley gave 66 interpretations in his book Enumerative Combinatorics 2, and he added some more as an addendum), you might be surprised to know that up to date, there is no known combinatorial interpretation of $S(m,n)$. However, there are some combinatorial interpretations of the specific case of super Catalan numbers in terms of pairs of Dyck paths with restricted height (see a paper by Gessel, for instance), cubic plane trees (see a paper by Pippenger and Schleich), or restricted lattice paths (see this paper by Chen and Wang). The first two papers give a combinatorial interpretation for $\dfrac{1}{2}S(m,2)$ and the last one gives a combinatorial interpretation for $S(m,m+s)$ where $0 \le s \le 4$.

There are also weighted interpretations of super Catalan numbers in terms of Krawtchouk polynomials in this paper by Georgiadis, Munemasa, and Tanaka. The most recent work on super Catalan numbers is due to Allen and Gheorghiciuc, who provided a weighted interpretation in terms of positive and negative 2-Motzkin paths in their paper here.

My current research is closely connected to super Catalan numbers, but not in a combinatorial flavor. However, I was curious why people haven’t been able to interpret these seemingly harmless numbers as a counting problem. To do that, I decided to do a bit of detour: I wandered around in the realm of combinatorics. Since one of the interpretations of the Catalan numbers is in terms of Dyck paths, I decided to start with Dyck paths.

To those who are not familiar, a Dyck path $D \in \mathcal{D}_n$ of length $2n$ is a lattice path, starting at $(0,0)$ and ending at $(2n,0)$, that consists of up-steps $(1,1)$ and down-steps $(1,-1)$ and never goes below the $x$-axis. You could also think of Dyck paths of length $2n$ as such paths starting from $(0,0)$ to $(n,n)$ that lies below (but may touch) the diagonal line $y = x$. The number of Dyck paths of length $2n$ is the $n$th Catalan number $\mathcal{C}_n$. You can see some illustrations here.

It thus makes sense to find a combinatorial interpretation of super Catalan numbers in terms of Dyck paths, hoping that the description of $S(m,n)$ relies solely on Dyck paths, only perhaps in more intricate detail. In order to come up with an interpretation, all I need to do is to find a very specific bijection between two Dyck paths that should work in general. That turns out to be a very difficult thing to do. I played with many bijections but none seemed to work. I ended up finding a curious family of bijections between Dyck paths of the same length, though. But more on that later, in another post.

So the moral of the story is you can wish to solve something, but you may end up getting something else. If you are lucky (or unlucky, depends on how you interpret it), that ‘something else’ might just be another question waiting to be discovered 😛