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 $k$?

  • 5
    $\begingroup$ Let $H=(\binom{2i+2j+1}{i+j})$ and $A=(\binom{2i+1}{i-j})$. Then $H=AA^t$ and therefore $detH=(detA)^2=1.$ $\endgroup$ Mar 24, 2018 at 20:33
  • 5
    $\begingroup$ Have you looked through Krattenthaler's determinant calculus papers? $\endgroup$ Mar 25, 2018 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. $\endgroup$ Mar 25, 2018 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 non-central binomial coefficients, and might produce a representation as a Gram matrix with respect to the inner product $\langle f, g \rangle = \int_{0}^{\infty} f(x) g(x) dx$. $\endgroup$ Mar 26, 2018 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, 2018 at 14:36

2 Answers 2


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}

  • $\begingroup$ 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? $\endgroup$ Jun 5, 2018 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 (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.

$$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n}=(2n+1)^{k}.$$ $$\det\left(\binom{2i+2j+2k+1}{i+j}\right)_{i,j=0}^{(2k+1)n-k-1}=(-1)^{\binom{k+1}2}(4n)^{k}.$$

  • $\begingroup$ 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? $\endgroup$ Apr 23, 2018 at 11:40
  • $\begingroup$ Thanks. I have included a link above. If this is still not clear, please do let me know. $\endgroup$ Apr 23, 2018 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? $\endgroup$ Apr 24, 2018 at 11:12
  • $\begingroup$ Other variant: $$\det\left(\binom{2i+2j+2k}{i+j}\right)_{i,j=0}^{k(2n+1)}=(-1)^{\binom{k}2+kn+1}[4k(n+1)]^{k-1}.$$ $\endgroup$
    – Wolfgang
    Apr 24, 2018 at 15:53
  • $\begingroup$ Oh, and also $$\det\left(\binom{2i+2j+2k}{i+j}\right)_{i,j=0}^{k(2n-1)-1}=(-1)^{\binom{k+1}2+kn}(4kn)^{k-1}.$$ $\endgroup$
    – Wolfgang
    Apr 26, 2018 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.