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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.)

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  • $\begingroup$ Since you already know they are proportional, a possible approach would be to compute $D$ at some very special set of values of the $a$'s, like $a_i=i$ or something like that $\endgroup$
    – Marcel
    Mar 4 at 20:01
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    $\begingroup$ I believe $c(8)=114397920$. This number has 97 as a prime factor, which suggests a very simple formula for $c(n)$ is unlikely $\endgroup$
    – Marcel
    Mar 4 at 20:09
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    $\begingroup$ In fact, $\sum_\sigma \mathrm{sgn}(\sigma) f(x_\sigma)$ is an alternating polynomial for any polynomial $f$. Hence, it is always a multiple of Vandermonde polynomial $V(x)$, and their ratio in general is a symmetric polynomial. $\endgroup$ Jul 28 at 15:33
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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 $c(n)$.

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    $\begingroup$ alternatively, $c(n)=[x_1^{n-1}x_2^{n-1}\ldots x_n^{n-1}] V(x_n,x_{n-1},\ldots,x_1)V(x_1,x_2-x_1,\ldots,x_n-x_{n-1})$, where $V(x_1,\ldots,x_n)=\prod_{i<j} (x_j-x_i)$. $\endgroup$ Jul 27 at 5:07
  • $\begingroup$ @Alekseyev. Yes, I agree. Maybe you can add a one-sentence explanation (e.g., this follows from looking at the coefficient of $a^{(0,1,...,n-1)}$ in each summand in the definition of $D$.) But still, is $c(n)$ non-zero for all $n$? $\endgroup$
    – Fan Ge
    Jul 27 at 19:48
  • $\begingroup$ @Petrov. Neat expression! $\endgroup$
    – Fan Ge
    Jul 27 at 19:53
  • $\begingroup$ @FanGe: 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. The question on whether $c(n)\ne 0$ seems to be hard. $\endgroup$ Jul 27 at 20:41
  • $\begingroup$ @Alekseyev. I wasn't asking why but was suggesting a clarification in your answer. But anyway, very nice observation! $\endgroup$
    – Fan Ge
    Jul 27 at 22:38

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