Ira Gessel "dubbed" the name *super Catalan* to
$$S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}$$
and offers a combinatorial proof in [his paper](http://people.brandeis.edu/~gessel/homepage/papers/superballot.pdf) 


**Note.** The numbers $\frac12S(m,n)$ are also integers.


I would like to extend the discussion by asking for a proof that the following numbers, which I call *super super Catalan numbers type 1*, are integers
$$S(x,y,z)=\frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!}$$
and the same question of integrality about the numbers, which I call *super super Catalan numbers of type 2*, 
$$T(x,y,z)=\frac{x}3\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3(x+y+z)}$$
provided that $x, y, z$ are non-negative integers (plus $x+y+z>0$ for the latter).


**Remark.** It is evident that $\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3}$ are integral; the extra factor $x$ in the numerator of $T(x,y,z)$ is necessary, although $y$ or $z$ would do.


I don't have a proof to these claims, but I am convinced of their truth. Even a generating function method is acceptable, yet a combinatorial proof is much desirable.



Extending the discussion on "super Catalan" numbers
I have a somewhat more algebraic proof than Jan-Christoph Schlage-Puchta's carry-based proof. I believe the idea goes back to Landau.


Let's start with your super super Catalan numbers type 1. A rational number $a$ is integer iff $v\_p(a)\ge 0$ for all primes $p$. Hence, to prove integrality of those numbers, it suffices to show that
$$(\*) \forall p: v\_p( \frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!} ) \ge 0 .$$
Define the following function $f:\mathbb{R}^3 \to \mathbb{Z}$:
$$ f(x\_1,x\_2,x\_3) = \lfloor 3x\_1 \rfloor + \lfloor 3x\_2 \rfloor + \lfloor 3x\_3 \rfloor - 2\left( \lfloor x\_1 \rfloor + \lfloor x\_2 \rfloor + \lfloor x\_3 \rfloor \right) - \lfloor x\_1+x\_2+x\_3 \rfloor.$$
Legendre has shown that
$$v\_p(n!) = \sum\_{ k \ge1} \lfloor \frac{n}{p^k} \rfloor,$$
which implies that
$$(\*\*) v\_p( \frac13\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!} ) = \sum\_{k \ge 1} f(\frac{x}{p^k}, \frac{y}{p^k}, \frac{z}{p^k}) - \delta\_{p=3}.$$
Hence, to establish $(\*)$, at least for $p \neq 3$, it suffices to show that $f$ is always non-negative. Note that $f$ has period 1 in each of its variables. Hence, WLOG we may assume that $0\le x\_i <1$, in which case
$$(\*\*\*) f(x\_1,x\_2,x\_3) = \lfloor 3x\_1 \rfloor + \lfloor 3x\_2 \rfloor + \lfloor 3x\_3 \rfloor - \lfloor x\_1+x\_2+x\_3 \rfloor.$$
Write $x\_i$ as $\frac{a\_i}{3}+r\_i$, where $a\_i \in \{0,1,2\}$ and $0 \le r\_i < \frac{1}{3}$. Then $(\*\*\*)$ becomes
$$f(x\_1,x\_2,x\_3) = a\_1+a\_2+a\_3 - \lfloor \frac{a\_1+a\_2+a\_3}{3} + (r\_1+r\_2+r\_3) \rfloor,$$
which is non-negative since
$$\frac{a\_1+a\_2+a\_3}{3} + (r\_1+r\_2+r\_3) < \frac{a\_1+a\_2+a\_3}{3} + 1 \le a\_1+a\_2+a\_3 + 1.$$




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What about $p=3$? Because of the extra term $-\delta\_{p=3}=-1$ in $(\*\*)$, we must show that at least one of the summands in $(\*\*)$ is positive. If we let $k$ equal the largest integer such that $3^{k-1} \mid x,y,z$, and write the fractional values of $\frac{x}{3^k}, \frac{y}{3^k}, \frac{z}{3^k}$ as $\frac{a\_1}{3} + r\_1, \frac{a\_2}{3}+r\_2, \frac{a\_3}{3} + r\_3$, then:
$$r\_1=r\_2=r\_3=0, a\_i \in \{0,1,2\}, a\_1+a\_2+a\_3 \ge 1.$$
Hence, the calculations above imply that 
$$f(\frac{x}{3^k}, \frac{y}{3^k}, \frac{z}{3^k}) = a\_1+a\_2+a\_3 -\lfloor \frac{a\_1 + a\_2+a\_3}{3} \rfloor \ge 1, $$ which implies that the $k$'th summand in $(\*\*)$ is positive.




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As for the super super Catalan numbers of type 2, note that
$$T(x,y,z) = S(x,y,z) \binom{x+y+z-1}{x-1,y,z},$$
so it is integer as a product of two integers.




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**Remark:** The same method shows that $C\_m(a\_1,\cdots,a\_n)=\frac{\prod\_{i=1}^{n} (m \cdot a\_i)!}{\prod\_{i=1}^{n} a\_i!^{m-1} (\sum a\_i)!}$ is integer for all positive integers $a\_1,\cdots,a\_n,m$, and that $\frac{C\_m}{m}$ is integer whenever $m$ is prime. Although I'm not sure that this is the "right" generalization of the super Catalan numbers.


