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My question is a bit imprecise - but I hope you like it. I even strongly think it has a proper answer.

The binomial coefficient $\binom{\frac{1}{2}}{n}$ is strongly related to Catalan numbers - the expression $(1-4x)^{\frac{1}{2}}$ appears when calculating the generating function of the Catalan numbers and solving a quadratic equation.

I am trying to find some combinatorial interpretation of $\binom{\frac{1}{k}}{n}$ for non-zero integer $k$. I feel it must exist - I don't know if it is because of intuition or because I've seen something similar and forgot. I want an elementary interpretation, maybe related to trees (since Catalan numbers count binary trees).

So, can anyone find a combinatorial interpretation of those coefficients (possibly multiplied by some power such as $k^n$)?

My motivation: I can show p-adically that $\binom{\frac{1}{k}}{n}$ is $p$-integral for any prime $p$ not dividing $k$. I am looking for a combinatorial proof of this property, and a combinatorial interpretation of $k^m \binom{\frac{1}{k}}{n}$ (for some integer $m$) will suffice for this.

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    $\begingroup$ $$ (-1)^{n-1} k^n \binom{\frac{1}{k}}{n} = \frac{1}{n!} \prod_{m=0}^{n-1} \left(m k - 1\right) $$ It does not look like $k^n \binom{1/k}{n}$ is integral for all $k$ and $n$. In fact for $k=n=3$ the answer is $\frac{5}{3}$. $\endgroup$ – Sasha Jul 17 '12 at 18:28
  • $\begingroup$ @Sasha You're right. Maybe $k^{2n} \binom{\frac{1}{2}}{n}$ is always integral? (Because in Catalan's numbers, the expression $2^{2n} \binom{\frac{1}{2}}{n}$ appears). $\endgroup$ – Ofir Jul 17 '12 at 19:38
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Propp's Exponentiation and Euler Measure contains such an interpretation.

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  • $\begingroup$ Looks interesting, I will read it. Thanks. $\endgroup$ – Ofir Jul 17 '12 at 19:47

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