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Consider $n\times n$ symmetric Cauchy-like matrix $M$ with elements $(M_{ij})_{i,j=1}^{n}$ given by

$$M_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x.$$

Is there a way to compute the elements of the inverse $(M^{-1})_{ij}$ analytically?

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    $\begingroup$ I don't have a proof, but empirically it seems that $\det(M) = 1/a(n)$ where $a$ is OEIS sequence A107254, and that the matrix elements of $M^{-1}$ are all integers. $\endgroup$ – Robert Israel Aug 26 '20 at 2:49
  • $\begingroup$ @RobertIsrael: I can prove the determinant formula you wrote. I am not sure if computing the adjugate is the cleanest way to proceed for inverse. $\endgroup$ – Abhishek Halder Aug 26 '20 at 4:13
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I was able to figure this out by viewing $M$ as a scaled Cauchy matrix.

Theorem. $\left(M^{-1}\right)_{ij} = \dfrac{(n-i)!(n-j)!}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)}$.

Proof. Define $n\times 1$ vector $\alpha$ with elements $(\alpha)_{i} = 1/(n-i)!$. Then $M = \text{diag}(\alpha) N \text{diag}(\alpha)$, where $N_{ij} := 1/(2n-i-j+1)$. Hence $\left(M^{-1}\right)_{ij} = (n-i)!(n-j)!\left(N^{-1}\right)_{ij}$.

Now, write $N$ as a Cauchy matrix: $N_{ij} = 1/(a_i + b_j)$ where $a_i := n-i$, $b_j := n-j+1$. Then using the known result [1, Sec. 1.2.3, Exercise 41] for Cauchy matrix inverse: $$\left(N^{-1}\right)_{ij} = \dfrac{1}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)},$$ the result follows.

[1] D. E. Knuth, The Art of Computer Programming. Volume 1: Fundamental Algorithms, 3rd ed. Addison- Wesley, 1997.

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