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][1] 

**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.

[1]:http://people.brandeis.edu/~gessel/homepage/papers/superballot.pdf