## A generalization of Catalan numbers

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?

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With sage can't find other (p,k) even with the relaxation the first few terms to be nonintegral. (12,1) and (18,1) are very rarely nonintegral though. – joro Sep 10 at 11:30

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.

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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:

http://arxiv.org/abs/math/9811086

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See also math.harvard.edu/~elkies/Misc/catalan.pdf and math.harvard.edu/~elkies/Misc/catalan2.pdf for elementary proofs of the formula for the relevant generating function. – Noam D. Elkies Sep 10 at 13:41
I accepted the other answer because it answered directly the question which I had for years - but this one is also very good! – Felix Goldberg Sep 10 at 16:41