It is well known that for $n>0$
$$d(n)=\det\left(\binom{2i+2j+1}{i+j}\right)\_{i,j=0}^{n-1}=1.$$
Computer experiments suggest that more generally
$$d(n,k)=\det\left(\binom{2i+2j+2k+1}{i+j}\right)\_{i,j=0}^{(2k+1)n-1}=(2n+1)^{k}.$$
Has anyone an idea how to prove this for general $k$?



Some binomial coefficient determinants
Johann Cigler and I have posted a solution on arXiv:


["An interesting class of Hankel determinants"](https://arxiv.org/abs/1807.08330), arXiv:1807.08330.


Let $d\_r(N)=\det\left({2i+2j+r\choose i+j}\right)\_{i,j=0}^{N-1}$. We show that for $k,n\ge 1$,
\begin{align}
&d\_{2k+1}((2k+1)n)=d\_{2k+1}((2k+1)n+1)=(2n+1)^k,\\
&d\_{2k+1}((2k+1)n+k+1)=(-1)^{k+1\choose 2}4^k(n+1)^k,\\
&d\_{2k}(2kn)=d\_{2k}(2kn+1)=(-1)^{kn},\\
&d\_{2k}(2kn+k)=-d\_{2k}(2kn+k+1)=(-1)^{kn+{k\choose 2}}4^{k-1}(n+1)^{k-1}.
\end{align}


