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Let $i,$ $j,$ $k$ be nonnegative integers such that $i+j+k$ is even. The expression $$(-1)^{j+k}\binom{i+j+k}{i,j,k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}$$ apparently computes the coefficients of certain terms in the expansion of a certain polynomial. (My answer to this question provides motivation. Incidentally, that question has an imminently-expiring bounty and deserves more attention - certainly something better than my own poor contribution.) It can be shown - see below - that this expression is invariant under permutation of $i,$ $j,$ $k.$ My question is, is there a natural way of rewriting the expression so that this symmetry is more apparent? Since the multinomial coefficient is manifestly symmetric, the main focus is on the remaining factors, $$g(i,j,k):=(-1)^{j+k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}.$$ If $g(i,j,k)$ equals, or is closely related to some named object, that would be very interesting to me as well.

The restriction that $i+j+k$ be even is necessary: in the case $i+j+k$ odd, $g(i,j,k)$ is, in fact, indeterminate for $k>i,j$; the sign is also changed by certain permutations in that case.

Some thoughts: We can express $g(i,j,k)$ in terms of generalized binomial coefficients as $$g(i,j,k)=(-1)^{j+k}\frac{\binom{\frac{1}{2}(i-j+k-1)}{k}}{\binom{\frac{1}{2}(i+j+k-1)}{k}}=(-1)^j\frac{\binom{\frac{1}{2}(-i+j+k-1)}{k}}{\binom{\frac{1}{2}(i+j+k-1)}{k}},$$ but this does not seem helpful.

To prove the symmetry it suffices to show that $g(i,j,k)=g(j,i,k)=g(k,j,i)$ since these two permutations generate the full permutation group. The first equality is proved as follows: $$\begin{aligned}g(j,i,k)&=(-1)^{i+k}\prod_{\ell=0}^{k-1} \frac{-i+j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=(-1)^{i+k}\prod_{\ell=0}^{k-1} \frac{-i+j-k+2\ell+1}{i+j+k-2\ell-1}\\ &=(-1)^i\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\&=g(i,j,k).\end{aligned}$$ The second line is obtained from the first by substituting $k-1-\ell$ for $\ell$ in the numerator; the last line follows since $(-1)^i=(-1)^{j+k}$ by the assumption $i+j+k$ even.

Equality of $g(i,j,k)$ and $g(k,j,i)$ is proved by taking $k\ge i$. We have $$\begin{aligned}1&=\prod_{\ell=0}^{k-i-1}\frac{i-j-k+2\ell+1}{i-j-k+2\ell+1}\\ &=\prod_{\ell=0}^{k-i-1}\frac{-i-j+k-2\ell-1}{i-j-k+2\ell+1}\\ &=\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{-i-j-k+2\ell+1}\\ &=(-1)^{k-i}\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}.\end{aligned}$$ The second line is obtained by substituting $k-i-1-\ell$ for $\ell$ in the numerator; the third is obtained by substituting $\ell-i$ for $\ell$ in both numerator and denominator. Then $$\begin{aligned}g(k,j,i)&=(-1)^{i+j}\prod_{\ell=0}^{i-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=(-1)^{i+j}\prod_{\ell=0}^{i-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}(-1)^{k-i}\prod_{\ell=i}^{k-1}\frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}\\ &=g(i,j,k).\\ \end{aligned}$$

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The formula looks reminiscent of the McMahon formula for counting plane partitions (although it is genuinely different), which is also a function of three variables invariant under permutations of the variables. However in that case the invariance is clear either from the formula or the properties of plane partitions. Perhaps there are ways of rewriting the McMahon formula that carry over to your example and make it look more symmetric. Alternatively, it would also be nice to say that these values count objects such that the number of objects is clearly invariant under permuting $i,j,k$. – Matthew Pressland Mar 13 '13 at 22:01
@MattPressland Thanks for pointing out the similarity with the McMahon formula. As you say, the symmetry of the McMahon formula is is present in the formula itself, and not just in the numbers it computes. But the connection bears further investigtion. – Will Orrick Mar 14 '13 at 1:18
In the case of my formula, it is easy to see that the coefficients it computes do possess the symmetry. But since I do not yet have proof of my formula, and because my formula is not manifestly symmetric, I had feared that perhaps it would turn out to hold only for restricted values of $i,$ $j,$ $k$, say for $i\ge j\ge k,$ and that the symmetry would have to be imposed "by hand". It was a pleasant surprise to find that no, the formula works for arbitrary $i,$ $j,$ $k,$ and the symmetry is automatically present. – Will Orrick Mar 14 '13 at 1:23
up vote 5 down vote accepted

Let $\mathbb O$ be the set of all odd integers. Define a function $Q:\mathbb O\to\mathbb Q$ by requiring

$Q\left(2n-1\right) = \left\lbrace \begin{array}{c} \prod\limits_{i=1}^{n}\left(2i-1\right),\text{ if }n > 0; \\ \prod\limits_{i=n+1}^{0}\dfrac{1}{2i-1},\text{ if }n \leq 0 \end{array} \right.$ for every $n\in\mathbb Z$.

Thus, every $m\in\mathbb O$ satisfies $Q\left(m\right) = m Q\left(m-2\right)$, and we have $Q\left(-1\right)=Q\left(1\right)=1$.

(Note how, unlike the factorial function, $Q$ is well-defined both on the positive and the negative side.)


$g\left(i,j,k\right) = \left(-1\right)^{\left(i+j+k\right)/2}\dfrac{Q\left(i+j-k-1\right)Q\left(j+k-i-1\right)Q\left(k+i-j-1\right)}{Q\left(i+j+k-1\right)}$.

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