Return to Question

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 $n$?

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.)

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 the total degree of $D$ (as a multivariable polynomial) is less than or equal to that of $V$, 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 $n$?

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 $n$?

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 degree, 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 $n$?

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 the total degree of $D$ (as a multivariable polynomial) is less than or equal to that of $V$, 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 $n$?