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Let $A_{n,m}$ be the band matrix $$ A_{n,m}=\left( \binom{2m}{j-i+m}-\binom{2m}{m-i-j-1} \right)_{0\leq {i,j} \leq {n-1}}.$$ For example,

$$A_{6,2}=\left ( \begin{matrix} 2 & 3 & 1& 0 & 0 & 0 \\ 3 & 6 & 4 &1 & 0 & 0\\1 & 4 & 6 &4 & 1&0 \\0 & 1 & 4 & 6 & 4 & 1\\ 0 & 0 & 1 & 4 & 6 & 4\\0& 0 & 0 &1 &4 & 6 \end{matrix} \right ).$$

As a corollary of the well-known formula $\det{\left( C_{i+j+m} \right)} _{0 \leq {i,j} \leq {n-1}}=\prod_{1 \leq i \leq j \leq {m-1}}{\frac{2n+i+j}{i+j}}$ for the Hankel determinants of the Catalan numbers it can be deduced that also $$\det A_{n,m} = \prod_{1 \leq i \leq j \leq {m-1}}{\frac{2n+i+j}{i+j}}.$$ Is there also a simple direct proof of this result?

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  • $\begingroup$ I have new post at mathoverflow.net/questions/447825/…, in case it makes your problem more amenable. $\endgroup$
    – T. Amdeberhan
    Commented May 29, 2023 at 19:12
  • $\begingroup$ Can you give a reference to the fact that "$\det A_{n,m}$ can be deduced from $\det(C_{i+j+m})$"? $\endgroup$
    – T. Amdeberhan
    Commented Jun 22, 2023 at 20:28
  • $\begingroup$ 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 $B_{n,m}=B_{n,0}A_{n,m}.$ $\endgroup$
    – Johann Cigler
    Commented Jun 23, 2023 at 14:32
  • $\begingroup$ Many thanks, indeed! $\endgroup$
    – T. Amdeberhan
    Commented Jun 25, 2023 at 19:23

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