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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