TITLE: a Vandermonde-type of determinants summed over permutations
QUESTION [15 upvotes]: Let $S_n$ be the symmetric group. Consider
$$D:=\sum_{\sigma\in S_n} \text{sgn}(\sigma)\cdot \det\begin{pmatrix}1 & a_{\sigma(1)}-0 & (a_{\sigma(1)}-0)^2 & \cdots & (a_{\sigma(1)}-0)^{n-1} \\1 & a_{\sigma(2)}-a_{\sigma(1)} & (a_{\sigma(2)}-a_{\sigma(1)})^2 & \cdots & (a_{\sigma(2)}-a_{\sigma(1)})^{n-1} \\ \vdots & & \vdots & & \vdots \\1 & a_{\sigma(n)}-a_{\sigma(n-1)} & (a_{\sigma(n)}-a_{\sigma(n-1)})^2 & \cdots & (a_{\sigma(n)}-a_{\sigma(n-1)})^{n-1} \end{pmatrix} .$$
If $a_i=a_j$ then it is easy to see $D=0$. Thus, the Vandermonde determinant $V:=\prod_{1\le i<j\le n}  (a_j-a_i)$ divides $D$. Since $D$ and $V$ have the same total degree (or $D$ is the zero polynomial), they differ by a constant factor, say $c(n)=D/V$.
I used Mathematica and found
$$c(2)=3, \quad c(3)=6, \quad c(4)=10, \quad c(5)=-1260, \quad c(6)=-28224, \quad c(7)=-352800.$$
Question:

Is there a closed form formula for $c(n)$?
Is $c(n)$ non-zero for all $n$?

(The question is motivated by a conjecture of Sun.)

REPLY [2 votes]: I do not have a concise formula for $c(n)$, but it can be more or less easily computed from the expansion of
$$\prod_{1\leq i<j\leq n} (a_j - a_{j-1} - (a_i - a_{i-1})),$$
where we conveniently define $a_0 := 0$. Namely, we are concerned only about terms of multi-degree being some permutation of $\{0,1,\dots,n-1\}$, and $c(n)$ equals the sum of the coefficients of such terms multiplied by their multi-degree (permutation) signs.
(Notice that only such terms do not vanish under alternating summation over $\sigma$. Furthermore, the summation of each such term gives $V$ multiplied by the sign of the term multi-degree.)
For example, when $n=3$ the product yields the following terms of interest:
$$-3 a^{(2, 1, 0)} - 3 a^{(1, 2, 0)} -3 a^{(0, 2, 1)} - 2a^{(1, 0, 2)} + a^{(0, 1, 2)},$$
which give $$c(3) = (-3)(-1) + (-3)(+1) + (-3)(-1) + (-2)(-1) + 1(+1) = 6.$$
Here is a sample Sage code for computing $c(n)$.