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Jun 25, 2023 at 19:23 comment added T. Amdeberhan Many thanks, indeed!
Jun 23, 2023 at 14:32 comment added Johann Cigler Observing that $b(n,k)=\binom{2n}{n-k}-\binom{2n}{n-k-1}$ is the number of non-negative lattice paths from $(0,0)$ to $(2n,2k)$ we get $\sum_{k}b(m+i,k)b(j,k)=b(m+i+j,0)=C_{i+j+m}.$ Let $B_{n,m}=\left(b(i+m,j)\right)_{i,j=0}^{n-1}.$ Then $B_{n,m}B_{n,0}^T=\left( C_{m+i+j}\right) _{i,j=0}^{n-1}.$ The result follows from $B_{n,m}=B_{n,0}A_{n,m}.$
Jun 22, 2023 at 20:28 comment added T. Amdeberhan Can you give a reference to the fact that "$\det A_{n,m}$ can be deduced from $\det(C_{i+j+m})$"?
May 29, 2023 at 19:12 comment added T. Amdeberhan I have new post at mathoverflow.net/questions/447825/…, in case it makes your problem more amenable.
Apr 28, 2023 at 14:26 history asked Johann Cigler CC BY-SA 4.0