Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$x^k c(x)^{2k}=(c(x)-1)^k =\sum_{n\geq0}c(k,n)x^n.$$ Consider the determinants $$D(k,n,m)= \det\left(c(k,i+j+m+1)\right)_{0\leq i,j\leq n-1}$$ with $D(k,0,m)=1.$
Computations suggest that $$D(k,kn,m)=(-1)^{\binom{k}{2}n} (n+1)^m$$ for $m\in \{0,1,\dots,k\}.$
Edit: For example the determinants $[D(3,n,m)]_{0\leq n\leq12,0\leq m\leq3}$ are
$\begin{bmatrix}*1*&0&0&*-1*&0&0&*1*&0&0&*-1*&0&0&*1*\\*1*&0&-1&*-2*&0&2&*3*&0&-3&*-4&0&4&*5*\\*1*&1&-9&*-4*&-4&45&*9*&9&-126&*-16*&-16&270&*25*\\*1*&6&-69&*-8*&21&1044&*27*&-882&-6448&*-64*&6084&25830&*125*\end{bmatrix}$
Remark: If more generally $c(x,t)$ is the generating function of the Narayana polynomials and $(c(x,q)-1)^k =\sum_{n\geq0}c(k,n,q)x^n$, then it seems that the corresponding determinants are $$D(k,kn,m,q)=(-1)^{\binom{k}{2}n}q^{k^2 \binom{n}{2}} {[n+1]_q}^m$$ with $[n]_q=1+q+\dots+q^{n-1}.$
Any idea how to prove this (at least for $q=1$) ?
Further edit: Computations also suggest that for $2\leq m \leq k+2$ $$(-1)^{n \binom{k}{2}+\binom{k-1}{2}}D(k,k n+k-1,m)=P_{k,m}(n)$$ is a polynomial of degree $3(m-1)$ with leading term $$(-1)^m \frac{B_{2m-2}}{(2m-2)!}2^{2m-3} k^{2m-2}n^{3m-3}.$$ Is there any reason why the Bernoulli numbers $B_n$ appear in this situation?
