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Let $n, k$ be any positve integers. I'm wondering if the following determinant is well known $$D_{n,k}= \begin{vmatrix}1^k& 2^k & 3^k&\cdots & n^k \\2^k & 3^k & 4^k &\cdots& (n+1)^k \\ \vdots&\vdots&\vdots&\ddots&\vdots\\n^k & (n+1)^k & (n+2)^k &\cdots&(2n-1)^k \end{vmatrix} $$

One can also consider a more general case. Indeed. let $n, k$ be any positve integers, and $a, d\in\mathbb{R}$, then what's the value of the following $$D_{n,k,a, d}= \begin{vmatrix}a^k& (a+d)^k & (a+2d)^k&\cdots & (a+(n-1)d)^k \\(a+d)^k & (a+2d)^k & (a+3d)^k &\cdots& (a+nd)^k \\ \vdots&\vdots&\vdots&\ddots&\vdots\\(a+(n-1)d)^k & (a+nd)^k & (a+(n+1)d)^k &\cdots&(a+(2n-1)d)^k \end{vmatrix} $$

Thanks in advance.

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    $\begingroup$ What do you want to know about these determinants? Sticking to $D(n,k)$, we easily have $D_{n,k}=0$ if $n>k+1$, and there seems to be a nice factorization for $D_{k+1,k}$; also there have been several Mathoverflow questions and answers about the sign of $D_{n,k}$, even for $k \notin \bf Z$. On the other hand if you fix $n$ and vary $k$ it's some complicated signed sum of $n!$ or so $k$-th powers. $\endgroup$
    – Noam D. Elkies
    Commented Dec 31, 2019 at 3:34
  • $\begingroup$ @NoamD.Elkies. Thank you very much. I'm very interested in the case $n=k+1$, where I want to know if there is an explicit and simple formula, but I cannot find any similar questions in MO. $\endgroup$
    – Tomas
    Commented Jan 3, 2020 at 2:58

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