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I have a sequence of real anti-symmetric matrices $M(k),k=1,2,\dots$, where $M(k)$ is a $2k\times 2k$ matrix with $(i,j)$ th element defined as $$ M(k)_{ij} =\frac{i-j}{i+j},\,1\le i\le2k,\, \,1\le j\le2k $$

Using Mathematica, I calculated $f(k)=\sqrt {\displaystyle\frac{\det(M(k))}{\det(M(k+1))}}$ for first few matrices.

$$ f(1)=350,\, f(2)=58212, \,f(3)=11042460, \,f(4)=2245709180, \,f(5)=476899543848 $$

I was surprised to see whole numbers as answer. Can anyone please help me prove or disprove that $f(k)$ is a whole number for all k.

user10001
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1 Answers1

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Here $[a_{ij}]$ notation means matrix with elements $a_{ij}$.

Let $n = 2k$

$$ \det\left(\left[\frac{i-j}{i+j}\right]\right) = \det\left(\left[\frac{i-j}{i+j} - 1\right]\right) = \det\left(\left[\frac{-2j}{i+j}\right]\right) = 2^n n! \det\left(\left[\frac{1}{i+j}\right]\right). $$

First equivlence is proved in this answer. (Credits to @user10001 for pointing out in comments that it is not as trivial as I originally thought, and for finding the reference for the proof).

Matrix $\left[\frac{1}{i+j}\right]$ is closely related to Hilbert matrix. This answer tells that $\det\left(\left[\frac{1}{i+j}\right]\right)=\det H_n / \binom{2n}{n}$. We'll use formula from the wikipedia for the determinant of Hilbert matrix: $$ \det H_n = \frac{\left( \prod\limits_{i=1}^{n-1} i! \right)^4}{\prod\limits_{i=1}^{2n-1}i!}. $$

Putting all together we get: $$ \det M(k) = \frac{2^n n! \left( \prod\limits_{i=1}^{n-1} i! \right)^4 \cdot n! \cdot n!}{\prod\limits_{i=1}^{2n-1}i! \cdot (2n)!} = \frac{2^n \left( \prod\limits_{i=1}^{n} i! \right)^4}{n! \cdot \prod\limits_{i=1}^{2n}i!}. $$

Simplifying $M(k)/M(k+1)$ we get: $$ \frac{\det M(k)}{\det M(k+1)} = \frac{(n+1)(n+2)(2n+1)!(2n+2)!(2n+3)!(2n+4)!}{4((n+1)!(n+2)!)^4} \\= \frac14 \cdot \frac{(2n+1)!}{n!(n+1)!}\cdot \frac{(2n+2)!}{(n+1)!(n+1)!} \cdot \frac{(2n+3)!}{(n+1)!(n+2)!} \cdot \frac{(2n+4)!}{(n+2)!(n+2)!} = \frac 14 \binom{2n+1}{n} \binom{2n+2}{n+1} \binom{2n+3}{n+1} \binom{2n+4}{n+2}. $$

Notice that $\binom{2s}{s} = 2\binom{2s-1}{s-1}$ (proof), which allows us to further simplify the expression:

$$ \sqrt\frac{\det M(k)}{\det M(k+1)} = \binom{2n+1}{n} \binom{2n+3}{n+1}, $$

which is an integer.

D. Dmitriy
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    Thanks!. For completeness, let me add that the first equality in the first line of your proof uses a result proved here. – user10001 Aug 23 '22 at 04:20
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    Oops! That's a bit embarassing. I used that equality, because I thought it is "obviously" correct for general matrices, which of course is not true. I did double-check numerically some steps of the proof, so that gave me some confidence that derivations are correct, but turns out that equality was correct because of very different reasons (lucky me). Good find by you! – D. Dmitriy Aug 23 '22 at 07:23
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    @user10001, I've included the link you gave in the answer. – D. Dmitriy Aug 23 '22 at 07:36