Let $\Delta\_n(x\_1, \ldots, x\_n)$ denote the Vandermonde determinant $\displaystyle \prod\_{1 \leq i < j \leq n}(x\_j - x\_i)$. Let $c\_1, \ldots, c\_n$ and $K$ be nonnegative integers satisfying
$$c\_1 + \cdots +c\_n = K + m\frac{n(n-1)}{2},$$
where $m$ is a positive integer. Then the coefficient of $x\_1^{c\_1}\cdots x\_n^{c\_n}$ in the polynomial $(x\_1 + \cdots + x\_n)^{K}\Delta\_n(x\_1^m, \ldots, x\_n^m)$ is
\begin{equation}
\sum\_{\sigma \in S\_n}\mathrm{sgn}(\sigma)\frac{K!}{\prod\_{i=1}^{n}(c\_i - \sigma(i)m + m)!}.
\end{equation}
My questions are the following:


1. Is it possible to express this coefficient as product of terms? We know that this is possible for $m = 1$ (see Lemma $9.8$ in Tao and Vu: Additive Combinatorics on page $336$) and in the case of $c\_i = k-i$ for $i = 1, \ldots, n$ for any positive integer $m$ (see Lemma $9.9$ in Tao and Vu: Additive Combinatorics on page $336$).
2. If that is not possible in general for $m \geq 2$, for which choices of $c\_i$, is it possible (in particular for $m = 2$)?


Any help in this regard will be greatly appreciated. Thanks.



Coefficient of a term in a several variable polynomial multipled with Vandermonde determinant
Note that $\Delta\_n(x\_1^m,\ldots,x\_n^m)=\det(x\_i^{k\_j})$, where $k\_j:=(j-1)m$. Two key observations are the following (both work for arbitrary non-negative integers $k\_1<k\_2<\ldots<k\_n$, if $\sum c\_j=\sum k\_j+K$):


1. (algebraic) $$\left[x\_1^{c\_1}x\_2^{c\_2}\ldots x\_n^{c\_n}\right](x\_1+\ldots+x\_n)^K\det(x\_i^{k\_j})=\frac{(K-\sum k\_j)!}{\prod c\_j!}\det(c\_i^{\underline{k\_j}}),\quad\quad\quad(\heartsuit)$$
where we use Knuth's notation $x^{\underline{n}}=x(x-1)\ldots(x-n+1)$. To prove $(\heartsuit)$, replace the polynomial $F:=(x\_1+\ldots+x\_n)^K\det(x\_i^{k\_j})$ to $$\tilde{F}=
\left(x\_1+\ldots+x\_n-\sum k\_j\right)^{\underline{K}}\det(x\_i^{\underline{k\_j}})$$
(which has of course the same highest degree coefficients as $F$)  

and apply [this](https://mathoverflow.net/a/214930/4312) formula for the sets $A\_i=\{0,1,\ldots,c\_i\}$. Note that if $a\_i\in A\_i$, and $\tilde{F}(a\_1,\ldots,a\_n)\ne 0$, expanding the determinant we see that there must exist a permutation $\pi$ for which $a\_i(a\_i-1)\ldots (a\_i-k\_{\pi\_i}+1)\ne 0$, i.e., $a\_i\geqslant k\_{\pi\_i}$, in particular $\sum a\_j\geqslant \sum k\_j$. Further, $\sum a\_j-\sum k\_j$ is a non-negative integer which can not be equal to $0,1,\ldots,K-1$, thus $\sum a\_j\geqslant \sum k\_j+K=\sum c\_j$. But $a\_j\leqslant c\_j$ for all $j$, thus this all is possible only if $a\_j=c\_j$ for all $j$. Therefore, in the RHS of the cited formula there is (at most) unique non-zero term, and this immediately gives $(\heartsuit)$.


2. (combinatorial) If $c\_1<c\_2<\ldots<c\_n$ are non-negative integers and $\sum c\_j=\sum k\_j+K$, then both parts of $(\heartsuit)$, which we denote by $\theta(c\_1,\ldots,c\_n)$ are equal to the number of paths from the point $(k\_1,\ldots,k\_n)$ to the point $(c\_1,\ldots,c\_n)$, for which each of $K$ paths constitutes of increasing exactly one coordinate by 1, and all intermediate points belong to a domain $\{(t\_1,\ldots,t\_n):t\_1<t\_2<\ldots<t\_n\}$. This may be proved by induction on $K$ for fixed $k\_1,\ldots,k\_n$. The base case $K=0$ is clear. For the induction step, note that $\Theta$ satisfies a recurrence
$$
\Theta(c\_1,\ldots,c\_n)=\Theta(c\_1-1,c\_2,\ldots,c\_n)+
\Theta(c\_1,c\_2-1,\ldots,c\_n)+\ldots+\Theta(c\_1,c\_2,\ldots,c\_n-1),
$$
which corresponds to considering the summand taken from the last bracket $x\_1+\ldots+x\_n$. And in RHS we may remove all summands in which $\Theta$ has two equal arguments (say, if $c\_5-1=c\_4$, remove the fifth summand), because they are anyway equal to 0, as is seen from RHS of $(\heartsuit)$ (or from general properties of antisymmetric polynomials, if you prefer.) Now exactly the same recurrence is enjoyed by the number of paths (consider the last step). This proves the induction step.


So, your question reduces either to computing the determinants, or computing the number of such paths. In combinatorics, such paths are traditionally viewed as *skew standard Young tableaux*: consider the Young diagrams $\lambda$ with row lengths $c\_1\leqslant c\_2-1\leqslant c\_3-2\leqslant \ldots \leqslant c\_n-(n-1)$ and $\mu$ with row lengths $k\_1\leqslant k\_2-1\leqslant k\_3-2\leqslant \ldots \leqslant k\_n-(n-1)$. Then the path corresponds to getting $\mu$ from $\lambda$ be consecutively adding the boxes, with a condition that on each step we get a Young diagram again. If you write the numbers $1,2,\ldots,K$ consequently when adding the boxes, you get what is called a skew standard Young tableaux.


Now $(\heartsuit)$ reads as Feit's determinantal formula for the number of skew standard Young tableaux, case $m=1$, that is, $\lambda=\emptyset$, corresponds to some variant of a hook length formula, and the case $c\_j=\ell+j$ (up to permutation it is your second case) corresponds to counting skew standard Young tableaux of a shape which is a reverse Young diagram ($\mu$ is a rectangle), that is, again to hook length formula.


An interesting and rather enigmatic story is that there are other shapes for which the number of skew standard Young tableaux may be computed as a product formula. Some such cases were found in a series of papers by Igor Pak, Greta Panova and Alejandro Morales which you can find on Igor's [homepage](https://www.math.ucla.edu/%7Epak/papers/research.htm#s).


