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An $n\times n$ circulant matrix $\mathbf{X}_n$ has the form

\begin{align} \mathbf{X}_n= \begin{bmatrix} x_1 & x_2 & \cdots & x_{n-1} & x_n \\ x_2 & x_3 & \cdots & x_n& x_1 \\ \vdots & \vdots& \ddots & \vdots & \vdots \\ x_{n-1} &x_n & \cdots & x_{n-3} & x_{n-2} \\ x_n & x_1 & \cdots & x_{n-2} & x_{n-1} \\ \end{bmatrix}. \end{align} If $f(z)=x_1+x_2z+\dotsb+x_{n-1}z^{n-2}+x_nz^{n-1}$ and $\xi=e^{\frac{2\pi i}n}$ with $i=\sqrt{-1}$ then we've the determinant $$\det(\mathbf{X}_n)=\prod_{k=0}^{n-1}f(\xi^k).\tag1\label1$$ For example, $\det(\mathbf{X}_1)=x_1$, $\det(\mathbf{X}_2)=x_1^2-x_2^2$, $\det(\mathbf{X}_3)=x_1^3+ x_2^3 + x_3^3-3x_1x_2x_3$ and \begin{align} \det(\mathbf{X}_4) &=x_1^4-x_2^4+x_3^4-x_4^4-4x_1^2x_2x_4+4x_1x_2^2x_3+4x_1x_3x_4^2- 4x_2x_3^2x_4 \\ &\,\,\, \qquad -2x_1^2x_3^2+2x_2^2x_4^2.\end{align} I like to change this a little bit to the matrix $\widetilde{\mathbf{X}}_n$ given by \begin{align} \widetilde{\mathbf{X}}_n= \begin{bmatrix} x_1 & x_2 & \cdots & x_{n-1} & x_n \\ x_2^2 & x_3^2 & \cdots & x_n^2& x_1^2 \\ \vdots & \vdots& \ddots & \vdots & \vdots \\ x_{n-1}^{n-1} &x_n^{n-1} & \cdots & x_{n-3}^{n-1} & x_{n-2}^{n-1} \\ x_n^n & x_1^n & \cdots & x_{n-2}^n & x_{n-1}^n \\ \end{bmatrix}. \end{align} For example, $\det(\widetilde{\mathbf{X}}_1)=x_1$, $\det(\widetilde{\mathbf{X}}_2)=x_1^3-x_2^3$ and $$\det(\widetilde{\mathbf{X}}_3)= x_1^6+x_2^6+x_3^6-x_1^3x_2x_3^2-x_1^2x_2^3x_3-x_1x_2^2x_3^3.$$ In view of these and more examples, there seem to be some structure, so I would like to ask:

QUESTION. Is there a formula for $\det(\widetilde{\mathbf{X}}_n)$ similar to that of \eqref{1}?

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  • $\begingroup$ Not sure what one can say from three examples. What do you mean by more experimentation? What did it reveal? Do you expect a nice answer merely because of the circulant case? In fact, why stop at the Vandermonde determinant. Take alternants with exponents of the form $\lambda_i+n-i$. $\endgroup$
    – user61318
    Commented Jul 16, 2022 at 17:07
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    $\begingroup$ Sure that these are beautiful cyclic patterns. But if even for $n=3$, the determinant doesn't seem to allow any kind of decomposition and the discriminant is as it is, there won't be a lot of hope. Unless you can find "nice" eigenvectors, but even for e.g. $(x_1,x_2,x_3)=(1,2,4)$ there is nothing nice... :( $\endgroup$
    – Wolfgang
    Commented Jul 16, 2022 at 19:49
  • $\begingroup$ If I were tasked with understanding this, I'd use the UFD-ness of $\mathbb Z[x_1,\ldots,x_n]$ and/or $\mathbb C[x_1,\ldots,x_n]$, and make adroit choices of values of the $x$'s to get vanishing, ... much as we do for Vandermonde. $\endgroup$ Commented Jul 16, 2022 at 23:19

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