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In the last MO question , the following matrix is given: $$M_{ij}=\left[\frac{1+(-1)^{i+j}}{i+j-1}\right]$$ and its inverse has been discussed. Now the problem is further extended to a more general form: $$M'_{ij}=\frac{(z+1)^{i+j-1}-(z -1)^{i+j-1}}{i+j-1}$$ where $-1 \le z \le 1$.

Similarly, is there an explicit formula for the inverse of $M'$? How about if we remove the limitation on $z$ and assume $-\infty \lt z \lt \infty$?

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I will consider instead $a(n,z)=\frac{(z+1)^{n+1}-(z-1)^{n+1}}{2(n+1)}$ because then $a(0,z)=1.$

The Hankel determinants $\det\left(a(i+j,z)\right)_{i,j=0}^n=\frac{2^{n^2+n}}{ \prod_{j=1}^n (2j+1)\binom{2j}{j}^2}$ are independent of $z.$

Let now $$p(n,x)=\sum_{k=0}^n (-1)^k \binom{n}{k} x^{n-k} \sum_{j=0}^{\lfloor{\frac{k}{2}}\rfloor}(-1)^j z^{k-2j}\binom{k}{2j} \frac{(2j-1)!!(2n-2j-1)!!}{(2n-1)!!}.$$

The inverse of the Hankel matrix is $$(\left(a(i+j,z)\right)_{i,j=0}^n)^{-1}=\left(b(i,j,z)\right)_{i,j=0}^n,$$ where $b(i,j,z)$ is the coefficient of $x^iy^j$ in $\sum_{k=0}^n \frac{2k+1}{4^k}\binom{2k}{k}^2 p(k,x)p(k,y)$.

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