It is well known that for $m\in \mathbb N$ the Hankel determinants $$D_m(n)= \det\left(C_{i+j+m}\right)_{0\leq i,j\leq {n-1}}$$ satisfy $D_m(n)=p_m(n)$, where $p_m(n)=\prod_{1 \leq i \leq j \leq {m-1}}{\frac{2n+i+j}{i+j}}$ is a polynomial.
If we extend the Catalan numbers to $\mathbb Z $ by setting $C_n=0$ for $n<0,$ then $D_m(n)$ is defined for all $m \in \mathbb Z.$
Using Dodgson condensation it can be shown that for $m>0$ $$D_{-m}(n)=p_{m+1}(-n)$$ or equivalently that $D_m(n)=0$ for $0<n<m+1$ and $$D_m(n+m+1)=(-1)^\binom{m+1}{2}D_{m+1}(n).$$ Computer experiments suggest that this result can be generalized in the following way: Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$ c(x)^{k} =\sum_{n\geq0}C(k,n)x^n.$$ Extend $C(k,n)$ to negative $n$ by $C(k,n)=0$ for $n<0$ and consider the determinants $$D_{k,m}(n)= \det\left(C(k,i+j+m)\right)_{0\leq i,j\leq n-1}$$ for $m \in \mathbb Z$. Then for $m>0$ $$ D_{k,-m}(n)=(-1)^{\binom{m+1}{2}}D_{k,m-k+2}(n-m-1)$$ for $n\geq m+1.$
For example $(D_{6,-3}(n))_{n \geq 0}= (1,0,0,0, 1,0,-1,-2,0,2,3,0,-3,-4, \dots)$ with $(D_{6,-1}(n))_{n \geq 0}= (1,0,-1,-2,0,2,3,0,-3,-4, \dots).$
Any idea how to prove this?
