8
$\begingroup$

For $n=1,2,3,\ldots$ let $a_n$ denote the determinant $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$. Then $$a_1=0,\ a_2=-1,\ a_3=-17280,\ a_4= 1168415539200.$$

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

I even conjecture that $$a_n'=\frac{(-1)^{n(n-1)/2}a_n}{2\prod_{k=1}^n(k!(2k-1)!)}$$ is a positive integer for every integer $n>2$. Note that \begin{gather*}a_3'=1,\ a_4'=559,\ a_5'=10767500,\ a_6'=9372614611500. \end{gather*}

The question is similar to my previous question http://mathoverflow.net/questions/302130. But it seems that darij grinberg's method there does not work for the present question.

Any comments are welcome!

$\endgroup$
  • 1
    $\begingroup$ My method easily generalizes to showing that $\det\left(\left(x_i+y_j\right)^n\right)_{0\leq i \leq n-1,\ 0 \leq j\leq n-1}$ is divisible by $n^2 V\left(x_0, x_1, \ldots, x_{n-1}\right) V\left(y_0, y_1, \ldots, y_{n-1}\right)$ whenever $x_0, x_1, \ldots, x_{n-1}, y_0, y_1, \ldots, y_{n-1}$ are $2n$ integers and $n \geq 3$. (Here, $V\left(a_0, a_1, \ldots, a_{n-1}\right)$ means the Vandermonde determinant $\prod\limits_{0 \leq i < j \leq n-1} \left(a_j - a_i\right)$.) Also, ... $\endgroup$ – darij grinberg Jun 6 '18 at 7:17
  • 1
    $\begingroup$ ... we have $V\left(0^2, 1^2, \ldots, \left(n-1\right)^2\right) = \prod\limits_{k=1}^{n-1} \left(k!^2 \dbinom{2k-1}{k}\right) = \prod\limits_{k=1}^{n-1} \left(k \left(2k-1\right)!\right)$. Since you have two of these $V$ factors around, you can kill almost all of your denominator; only $2 n \left(2n-1\right)!$ remains. I'm not sure where to get the $\left(2n-1\right)!$ from, though, particularly if $2n-1$ is prime. $\endgroup$ – darij grinberg Jun 6 '18 at 7:23
4
$\begingroup$

The conjecture in my posting has been proved finally! Its solution will appear in a forthcoming joint paper by Darij Grinberg, Guoniu Han, Zhi-Wei Sun and Lilu Zhao.

$\endgroup$
1
$\begingroup$

This is not an answer but a claim regarding some generalized determinant: $$\det\left[(x_i+x_j)^n\right]_{i,j=0}^{n-1}$$ is divisible by the square of the Vandermonde $$\prod_{i<j}^{0,n-1}(x_i-x_j)^2.$$ Letting $x_i=i^2$ recovers your matrix.

$\endgroup$
  • $\begingroup$ It is symmetric polynomial and equals to 0 when two variables are equal. This implies your claim. $\endgroup$ – Fedor Petrov Jun 6 '18 at 5:57
  • $\begingroup$ Further, using the suggested substitution gives the square of the Vandermonde as the square of the product of the odd factorials up to 2n-1!. This comes close to resolving the posted conjecture. Gerhard "Lots Of Multiplication Going On" Paseman, 2018.06.05. $\endgroup$ – Gerhard Paseman Jun 6 '18 at 6:17
  • 1
    $\begingroup$ when we divide by the square of the Vandermonde we get a symmetric polynomial in $x_i$'s of degree $n$ which has degree 2 in each specific variable. Maybe it is explicitly computable? $\endgroup$ – Fedor Petrov Jun 6 '18 at 9:45
  • $\begingroup$ @FedorPetrov: Let $x_0, x_1, \ldots, x_{n-1}, y_0, y_1, \ldots, y_{n-1}$ be $2n$ elements of a commutative ring $R$. Let $\overrightarrow{x} = \left(x_0, x_1, \ldots, x_{n-1}\right)$ and $\overrightarrow{y} = \left(y_0, y_1, \ldots, y_{n-1}\right)$. For any $\overrightarrow{z} = \left(z_0, z_1, \ldots, z_{n-1}\right) \in R^n$, we let $V \left( \overrightarrow{z} \right)$ be the Vandermonde determinant $\prod\limits_{0 \leq i < j \leq n-1}\left(z_j - z_i\right)$, and we let $e_i\left(\overrightarrow{z}\right)$ be the $i$-th elementary symmetric polynomial in ... $\endgroup$ – darij grinberg Jun 6 '18 at 13:23
  • 1
    $\begingroup$ ... the $z_0, z_1, \ldots, z_{n-1}$ whenever $i$ is a nonnegative integer. (Note that $e_0\left(\overrightarrow{z}\right) = 1$.) Then, $\det \left( \left(x_i + y_j\right)^n \right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n-1}$ $= V\left(\overrightarrow{x}\right) V\left(\overrightarrow{y}\right) \sum\limits_{k=0}^n \left(\prod\limits_{i \in \left\{0,1,\ldots,n\right\};\ i \neq k} \dbinom{n}{i}\right) e_{n-k}\left(\overrightarrow{x}\right) e_k\left(\overrightarrow{y}\right)$. This follows from the argument using Cauchy-Binet that I gave in the comments to mathoverflow.net/questions/302130 . $\endgroup$ – darij grinberg Jun 6 '18 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.