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Based on my computation, I have formulated the following conjecture.

Conjecture. For any positive integer $n$, we have $$\det[|j-k|]_{1\le j,k\le n}=(-1)^{n-1}(n-1)2^{n-2}\tag{1}$$ and $$\det[|j^2-k^2|]_{1\le j,k\le n}=(-1)^{n-1}(2n-1)!!(n^2-1)2^{n-2}.\tag{2}$$

QUESTION. Are the identities $(1)$ and $(2)$ known? How to prove them?

This question looks not so difficult. Your comments are welcome!

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    $\begingroup$ After discussion with Prof. Bao-Xuan Zhu, we have got the following general formula: $$\det[|x_j-x_k|]_{1\le j,k\le n}=2^{n-2}(x_1-x_2)\cdots(x_{n-1}-x_{n})(x_n-x_1),$$ where $x_1<\ldots<x_n$. Both $(1)$ and $(2)$ follow from this general formula. $\endgroup$
    – Zhi-Wei Sun
    Commented Aug 28, 2021 at 9:12

2 Answers 2

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Identity (1) follows from Theorem 1 in the following paper:

Florian Bünger, Inverses, determinants, eigenvalues, and eigenvectors of real symmetric Toeplitz matrices with linearly increasing entries, Linear Algebra and its Applications 459 (2014), 95-619

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Regarding the (generalized) determinant commented by Zhi-Wei Sun, with $\pm$ correction, $$\det[\,\vert x_j-x_k\vert\,]_{1\leq j,k\leq n}=(-1)^{n-1}2^{n-2} (x_1-x_2)\cdots(x_{n-1}-x_n)(x_n-x_1),$$ the proof goes as follows: the LHS is a homogeneous multi-variable polynomial of degree $n$ which vanishes at consecutively-indexed variables (cyclically), so is the RHS. Furthermore, both sides are seen to be quadratic in each variable. It remains to compute the coefficient shown on the RHS. This, however, is determined if we set $x_j=j$ because the resulting determinant is precisely claimed in equation (1) and proven according to Thomas Kalinowski's reference.

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  • $\begingroup$ My formula with Bao-Xuan Zhu is correct, with no sign wrong. We had a proof when I wrote the comment yesterday. When $x_j=j$ for $j=1,\ldots,n$ it yields the identity $(1)$. In contrast, your modified version is wrong, you can see this if you take $x_j=j$ for $j=1,\ldots,n$. $\endgroup$
    – Zhi-Wei Sun
    Commented Aug 28, 2021 at 22:15

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