Question regarding determinants of a sequence of anti-symmetric matrices. 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.
 A: 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.
