Some binomial coefficient determinants 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 $k$?
 A: Let $C$ be the unit circle oriented positively, $z\in C$ and $\iota=\sqrt{-1}$. We know that
$$\binom{a}b=\frac1{2\pi\iota}\int_C\frac{(1+z)^a}{z^b}\frac{dz}z.$$
After an application of this formulation, the problem is equivalent to the multiple (Selberg-type) contour integral
$$\frac1{(2\pi\iota)^N}\int_C\cdots\int_C\prod_{j=1}^N\frac{(1+z_j)^{2k+2j-1}}{z_j^j}\prod_{j<m}^{1,N}\left(\frac1{z_m}+z_m-\frac1{z_j}-z_j\right)dz_1\cdots dz_N=(2n+1)^k;$$
where $N=(2k+1)n$. For additional information on such a transformation, you may look at this paper starting on page 3.

*
  Here are a couple of variants of the problem. Only the sizes of the matrices are altered.

$$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n}=(2n+1)^{k}.$$
$$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n-k-1}=(-1)^{\binom{k+1}2}(4n)^{k}.$$
A: Johann Cigler and I have posted a solution on arXiv:
"An interesting class of Hankel determinants", 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}
