It is well known that for $n>0$ $$d(n)=\det\left(\binom{2i+2j+1}{i+j}\right)_{i,j=0}^{n1}=1.$$ Computer experiments suggest that more generally $$d(n,k)=\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n1}=(2n+1)^{k}.$$ Has anyone an idea how to prove this for general $k$?

5$\begingroup$ Let $H=(\binom{2i+2j+1}{i+j})$ and $A=(\binom{2i+1}{ij})$. Then $H=AA^t$ and therefore $detH=(detA)^2=1.$ $\endgroup$ – Johann Cigler Mar 24 '18 at 20:33

5$\begingroup$ Have you looked through Krattenthaler's determinant calculus papers? $\endgroup$ – Steve Huntsman Mar 25 '18 at 0:36

5$\begingroup$ @PerAlexandersson it does not seem to be straightforward, even for $k=0$. We count the lattice paths (directions up und right) from the points $(i,i)$ to $(2k+1+j,j)$. But the order is not fixed and some cancellations happen. $\endgroup$ – Fedor Petrov Mar 25 '18 at 20:26

2$\begingroup$ One possible direction is to show that your $(2k+1)n$square matrix is a Gram matrix, and then there are several approaches to computing its determinant. We have the identity $\binom{2n}{n} = \frac{2^{2n+1}}{\pi}\int_{0}^{\infty} \frac{dx}{(1+x^2)^{n+1}}$ (attributed to Wallis), which might be generalizable in some way to noncentral binomial coefficients, and might produce a representation as a Gram matrix with respect to the inner product $\langle f, g \rangle = \int_{0}^{\infty} f(x) g(x) dx$. $\endgroup$ – Ofir Gorodetsky Mar 26 '18 at 9:19

7$\begingroup$ The case $k=1$ follows as a special case of (49) pg. 19 of arxiv.org/pdf/0804.0440.pdf  their techniques don't seem tractable for the general case. $\endgroup$ – Suvrit Mar 26 '18 at 14:36
Johann Cigler and I have posted a solution on arXiv:
"An interesting class of Hankel determinants", arXiv:1807.08330.
Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,j=0}^{N1}$. 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^{k1}(n+1)^{k1}. \end{align}

$\begingroup$ Sorry for my late response. I have seen your answer only today. It seems to be the sort of proof I am looking for. But I have many problems with the details. It would take too long to discuss these here. I would rather discuss my difficulties privately with you by email. Would you agree to this procedure? $\endgroup$ – Johann Cigler Jun 5 '18 at 11:30
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 (Selbergtype) contour integral $$\frac1{(2\pi\iota)^N}\int_C\cdots\int_C\prod_{j=1}^N\frac{(1+z_j)^{2k+2j1}}{z_j^j}\prod_{j<m}^{1,N}\left(\frac1{z_m}+z_m\frac1{z_j}z_j\right)dz_1\cdots dz_N=(2n+1)^k;$$ 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.
$$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n}=(2n+1)^{k}.$$ $$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)nk1}=(1)^{\binom{k+1}2}(4n)^{k}.$$

$\begingroup$ This looks very interesting, but I do not understand how you got this. Could you please give me some hints how to obtain this reformulation? $\endgroup$ – Johann Cigler Apr 23 '18 at 11:40

$\begingroup$ Thanks. I have included a link above. If this is still not clear, please do let me know. $\endgroup$ – T. Amdeberhan Apr 23 '18 at 12:34

$\begingroup$ This is a nice trick to reduce the Hankel determinant to an integral of Vandermonde's determinant. But does it help to compute the determinant? $\endgroup$ – Johann Cigler Apr 24 '18 at 11:12

$\begingroup$ Other variant: $$\det\left(\binom{2i+2j+2k}{i+j}\right)_{i,j=0}^{k(2n+1)}=(1)^{\binom{k}2+kn+1}[4k(n+1)]^{k1}.$$ $\endgroup$ – Wolfgang Apr 24 '18 at 15:53

$\begingroup$ Oh, and also $$\det\left(\binom{2i+2j+2k}{i+j}\right)_{i,j=0}^{k(2n1)1}=(1)^{\binom{k+1}2+kn}(4kn)^{k1}.$$ $\endgroup$ – Wolfgang Apr 26 '18 at 12:39