# Variant of binomial coefficients

I've recently come across a variant of the binomial polynomials, and I'm curious if anyone has seen these before. If so, I'd love a reference, a name, etc.

First recall the following. If z is a formal variable, then we can consider \binom{z}{k} as a polynomial in z by the standard formula: \binom{z}{k}= [z(z-1)...(z-k+1))]*[k!]-1.

Here's the variant I came across. Let a and k integers, where a divides k, and we write k=ab. Consider the polynomial

F(a,k)(z)=[z(z-a)(z-2a)...(z-k+a))][k(k-a)(k-2a)...(a)]-1.

Has anyone this before? Or anything similar?

Update: Jonah helpfully identifies a typo in the numerator of F(a,k)(z) which I've now corrected.

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Well, from a polynomial perspective they're essentially the same. Consider F<sub>(a,k)</sub>(az) and cancel the factors of a. – Qiaochu Yuan Oct 29 '09 at 15:41
Thanks for all of these great answers! I had missed the change of coordinates and hadn't heard of multifactorials before. – Daniel Erman Oct 29 '09 at 17:22

(For simplicity, you probably want the last term in the numerator to be z-k+a, right? That way F(1,k)(z)=\binom{z}{k}. I'll pretend that's what you meant.)

I haven't come across such polynomials, but they're easily expressed in terms of multifactorials. Namely,

F(a,k)(z)=z!(a)/[(z-k)!(a)k!(a)]

Note that this isn't an integer when a doesn't divide z. EDIT: (And, as Qiaochu points out, when a does divide z it's just regular ol' z-choose-k.)

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On the other hand, maybe what you want is a combinatorial interpretation. There is a way to express a related function as a binomial determinant which can be combinatorially interpreted using Gessel-Viennot, as described here.

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That's pretty awesome. I should probably get better at recognizing Gessel-Viennot applications when they pop up. – Jonah Ostroff Oct 29 '09 at 17:21

Let Bk(z) denote the usual binomial coefficient, suitably generalized to allow z to be a formal variable. Assuming that you intend to have the same number of factors in the numerator and denominator in your definition of F(a,k)(z), then via the substitution z = aw,

F(a,ab)(z) = Bb(w).

If 1/a is in your ring (still assuming that a divides k), then

F(a,k)(z) = Bk/a(z/a).

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