It is well-known that the $n$th Catalan number is equal to $(n+1)^{-1}\binom{2n}{n}$. A long time ago I had wondered what happens if you look at the sequence generated by $(n+k)^{-1}\binom{pn}{n}$ - for which $p,k$ is it integral?
I found no other integral-producing values except for $(p,k)=(2,1)$ but then I'm probably missing something. (Also, MATLAB, which is my main tool, quickly runs out of precision on this kind of computation).
So - is this known? Trivial, perhaps?
Yes, $(p,k)=(2,1)$ is the only case when the sequence is integral.
We always need $p\geq 2$. First let's pick a prime $q$ which is $\geq k$ and which divides one of the numbers in $\lbrace pk-1,pk-2,\dots,pk-k\rbrace$. This always exists for $k\geq 2$ by a famous result of Sylvester.
Now taking $n=q-k$, it is easy to check that $n+k$ does not divide $pn(pn-1)\cdots((p-1)n+1)$ and so $\frac{1}{n+k}\binom{pn}{n}$ is not an integer.
Maybe you are looking for the Fuss-Catalan numbers?
These are given by $\frac{1}{(s-1)n+1}\binom{sn}{n}$.
A starting point for the combinatorial significance of these numbers is:
You might also be interested in the rational Catalan numbers, defined for $(a,b)=1$ as $\mathrm{Cat}(a,b) := \frac{1}{a+b}\binom{a+b}{a}$.
See these slides by Drew Armstrong for their significance: http://www.math.miami.edu/~armstrong/RCCinDC.pdf
As this paper explains, they are a further generalization of Fuss-Catalan numbers: http://arxiv.org/abs/1305.7286
One more related oject is the Catalan's triangle with integer entries $$B(n,r)=\frac rn{2n\choose n-r}$$ (see also the free book Catalan Numbers With Applications - T Koshy (Oxford, 2009)
