I wish to find the coefficients of the Laurent expansion of the inverse of the Vandermonde determinant, that is, the Laurent expansion at 0 of
$$\prod\_{1\leq i<j \leq n}(x\_j-x\_i)^{-1}.$$


We can iteratively do a series expansion in $x\_1,x\_2,$ and so on,
(the full expansion depends on the order of expansion).


An answer I would be satisfied with is if 
I can say that the coefficient $c\_a(k)$ of
$x\_1^{a\_1+k}x\_1^{a\_2+k}\cdots x\_{n-1}^{a\_{n-1}+k} x\_n^{a\_n-k}$ 
in the Laurent expansion of
$$\prod\_{1\leq i<j \leq n}(x\_j-x\_i)^{-d}.$$
is a polynomial in $k,$ for every fixed positive integer $d$.


A paper stating this, or a short argument of this fact would be terrific.



Laurent expansion of inverse of vandermonde determinant
In your question, the exponent of $x\_2$ is $a\_2+k$. I am assuming that is a typo, and you intended $c\_a(k)$ to be the coefficient of 
$$x\_1^{a\_1+k} x\_2^{a\_2} x\_3^{a\_3} \cdots x\_{n-2}^{a\_{n-2}} x\_{n-1}^{a\_{n-1}} x\_n^{a\_n-k}.$$
If so, I can answer all of your questions.


It will be convenient to switch the sign in your Vandermonde determinant, so I'll set $\Delta = \prod\_{1 \leq i < j \leq n} (x\_j-x\_i)$. Let $\rho$ be the vector $(0,1,2,\ldots, n-1)$. So
$$\Delta = x^{\rho} \prod\_{1 \leq i < j \leq n} ( 1- x\_i/x\_j)$$
and
$$\Delta^{-1} = x^{-\rho} \prod\_{1 \leq i < j \leq n} \frac{1}{1-x\_i/x\_j} 
=x^{-\rho} \prod\_{1 \leq i < j \leq n} \sum\_{c\_{ij}=0}^{\infty} \left( \frac{x\_i}{x\_j} \right)^{c\_{ij}}.$$


So the coefficient of $x^a$ in $\Delta^{-1}$ is the number of ways to write $a+\rho$ in the form $\sum\_{1 \leq i < j \leq n} c\_{ij} (e\_i-e\_j)$, where $e\_k$ is the $k$-th basis vector and the $c\_{ij}$ are nonnegative integers. This is called the Kostant partition function of $a+\rho$; I'll denote it by $K(a+\rho)$.


If you do the same argument for $\Delta^{-d}$, you are counting the number of ways to write
$$a+\rho = \sum\_{1 \leq i < j \leq n} \sum\_{1 \leq k \leq d} c\_{ij}^k (e\_i-e\_j)$$
where $c\_{ij}^{k}$ are nonnegative integers. Let's call this $K^d(a+\rho)$.


As I will explain below, the values of $K^d(\beta)$ are piecewise polynomial, with domains of polynomiality being convex polyhedral cones. The line $a+\rho + k (e\_1-e\_n)$ will, for $k$ sufficiently large, eventually lie in just one cone, so it will be polynomial for $k$ sufficiently large. I'll discuss the switch between "eventually polynomial" and "polynomial" below.


The piecewise polynomiality of $K^d(\beta)$ is a special case of the following theorem of [Blakley](http://projecteuclid.org/euclid.dmj/1077375226); I like the presentation of it in this paper of [Sturmfels](http://www.sciencedirect.com/science/article/pii/0097316595900675)


Let $v\_1$, $v\_2$, ..., $v\_N$ be a finite list of vectors in $\mathbb{Z}^M$, all lying in an open half space. Let $\Lambda$ be the lattice generated by the $v\_i$. For $\lambda \in \Lambda$, let $c(\lambda)$ be the number of ways to write $\lambda$ as $\sum c\_r \lambda\_r$, with $c\_r \in \mathbb{Z}\_{\geq 0}$. Then can partition $\mathbb{R} \Lambda$ into finitely many convex polyhedral cones $K\_j$, so that $c$ is a quasi-polynomial on $\Lambda \cap K\_j$.


Here a function $\phi$ is called quasi-polynomial means that we can find a finite index sublattice $M$ of $\Lambda$ so that $\phi$ is polynomial on cosets of $\Lambda/M$. However, you don't have to worry about the word "quasi" because of a refinement of this result, also stated in Sturmfels' paper: Suppose that, whenever $v\_{i\_1}$, $v\_{i\_2}$, ..., $v\_{i\_D}$ is a $\mathbb{Q}$-basis for $\mathbb{Q} \Lambda$, then $v\_{i\_1}$, $v\_{i\_2}$, ..., $v\_{i\_D}$ is a $\mathbb{Z}$-basis for $\Lambda$. Then we get honest polynomials instead of quasi-polynomials. In this case, we say that the $v$'s are **unimodular**. The $v$'s in our setting are unimodular, so we get honest polynomials. See the top of page 305 in Sturmfels' paper for more on this.


For discussion of practical computation of these polynomials for the Kostant partition function, see [de Loera and Sturmfels](http://arxiv.org/abs/math/0104286) and [de Loera's website](https://www.math.ucdavis.edu/~deloera/kostant.html).




---


For notational simplicity, set $\beta = a+\rho$, so $\sum \beta\_i=0$. You asked for $K^d(\beta+k(e\_1-e\_n))$ to be polynomial, not just eventually polynomial, in $k$. In a dumb sense, this is false. If $\beta = (-10, 0,0,\cdots, 0, 10)$, then $K(\beta + k(e\_i-e\_j))$ is $0$ for $k < 10$, and then becomes $\geq 0$. However, I'm guessing you might have meant the following:



> 
> Suppose that $\beta$ is in the positive integer span of $e\_i-e\_j$ for $i<j$ (so that $x^{\beta-\rho}$ is an exponent in the Vandermonde inverse). Then $K^d(\beta+k (e\_1-e\_n))$ is polynomial in $k$ for $k \geq 0$.
> 
> 
> 


I expect this is **false** for $n=4$ and $d=1$, with $\beta$ of the form $a (e\_1-e\_2) + b (e\_2-e\_3) + c (e\_3-e\_4)$, and $b \gg a > c$. Take a look at Figure 1 in the de Loera-Sturmfels paper, where they draw the chamber complex for $n=4$. I am describing a point in the region they label $3$. For $k \gg 0$, the point $\beta + k (e\_1-e\_4)$ will be in the region they label $4$. The polynomials on regions $3$ and $4$ are different. Now, different polynomials can become equal when restricted to a line, and I haven't had time to check that this doesn't happen in your case, but I know of no reason it should happen. I'll try to do some examples this evening.


**UPDATE (algebra errors in the earlier counter-example now corrected)** I now have a specific counterexample. Let $n=4$ and let $\beta = (0,m,-m,0)$. I get that (for $k$, $m \geq 0$) 
$$K(k,m,-m,-k) = (k+1)(k+2)(2k+3)/6 - \min(0, (k-m+1)(k-m)(k-m-1)/6).$$ 
So, for $m \geq 2$, this is not polynomial in $k \geq 0$. 


