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For $n=1,2,3,\ldots$ let $a(n)$ denote the determinant $\det[(i+j)^n]_{0\le i,j\le n-1}$.

QUESTION: Is it true that $n^2\mid a(n)$ for all $n=3,4,\ldots$?

I even conjecture that $$b(n)=\frac{(-1)^{n(n-1)/2}a(n)}{(n-2)!n\prod_{k=1}^nk!}$$ is a positive integer for every integer $n>2$. Note that \begin{gather*}b(3)=4,\ b(4)=229,\ b(5)=89200,\ b(6)=336775500, \\ b(7)= 15858447494400,\ b(8)= 11358391301972951040. \end{gather*}

Any comments are welcome!

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  • $\begingroup$ Let $A$ be the $n \times \left(n+1\right)$-matrix $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n}$. Let $B$ be the $\left(n+1\right)\times n$-matrix $\left(j^{n-k}\right)_{0 \leq k \leq n, \ 0 \leq j \leq n-1}$. Then, $AB$ is your $n \times n$-matrix $\left(\left(i+j\right)^n\right)_{0 \leq i \leq n-1, \ 0 \leq j \leq n-1}$ (by the binomial formula). Thus, $a\left(n\right) = \det\left(AB\right)$ can be computed by the Cauchy-Binet formula; you get a sum of several products of a maximal minor of $A$ and a maximal minor of $B$. Moreover, ... $\endgroup$
    – darij grinberg
    Commented Jun 5, 2018 at 23:22
  • $\begingroup$ ... the $1$st and $n-1$st columns of $A$ are divisible by $n$, so that each maximal minor of $A$ that contains both of these columns is divisible by $n^2$. It remains to consider the remaining two maximal minors: the one that misses column $1$, and the one that misses column $n-1$. If $n \geq 5$, then these two minors contain both the $2$nd and the $3$rd column of $A$, and thus have a factor of $\dbinom{n}{2}\dbinom{n}{3}$, which is also divisible by $n$ (as a simple case distinction shows). So it remains to solve the $n < 5$ cases, which you have done in your post. $\endgroup$
    – darij grinberg
    Commented Jun 5, 2018 at 23:26
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    $\begingroup$ Minor correction to my first comment: Of course, $A$ should be $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq k \leq n}$, not $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n}$. $\endgroup$
    – darij grinberg
    Commented Jun 5, 2018 at 23:35
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    $\begingroup$ @darij grinberg Great! You are the first one who can answer one of my questions posted in Mathoverflow. It seems that your method does not work for my another similar question which I will pose soon. $\endgroup$
    – Zhi-Wei Sun
    Commented Jun 5, 2018 at 23:49
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    $\begingroup$ A simpler variant is: $\det[(x+i+j)^{n-1}]_{i,j=1}^n=(-1)^{\binom{n}2}(n-1)!^n$. Even more general: $\det[(x_i+x_j)^{n-1}]_{i,j=1}^n=(-1)^{\binom{n}2}\prod_{k=1}^nk^{2k-1-n}V(x_1,\dots,x_n)$ where $V$ is Vandermonde determinant. $\endgroup$
    – T. Amdeberhan
    Commented Jun 5, 2018 at 23:58

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