Let $$C_n=\frac{1}{2n+1}\binom{2n+1}{n}$$ be a Catalan number. It is well-known that $$(\sum_{n\ge{0}}C_n x^n)^k=\sum_{n\ge{0}}C(n,k)x^n$$ with $$C(n,k)=\frac{k}{2n+k}\binom{2n+k}{n}.$$ It is also known that the Hankel matrix $\left( {{C(i+j,2)}} \right)_{i,j = 0}^{n - 1}$ can be factored in the form $$\left( {{C(i+j,2)}} \right)_{i,j = 0}^{n - 1}=A_{n} A_{n}^T,$$ where

$$A_{n}=\left(\binom{2i+1}{i-j}\frac{2(j+1)}{i+j+2}\right) _{i,j = 0}^{n - 1}=\left(\binom{2i+1}{i-j}-\binom{2i+1}{i-j-1}\right) _{i,j = 0}^{n - 1}.$$

Computer experiments suggest that for $k\ge{1}$ $$\left( {{C(i+j,k+2)}} \right)_{i,j = 0}^{n - 1}=A_{n}G_{n,k} A_{n}^T$$ with $$G_{n,k}=\left({g(i,j,k)}\right) _{i,j = 0}^{n - 1},$$ where $$g(i,j,k)= \sum_{m={|i-j|-1}}^{i+j}\binom{k-1}{m}. $$ Is there a simple proof of this identity?