For $\{h_{n}\}_{n=0}^{\infty}$ a real sequence, denote by $H_{n}$ the $n\times n$ Hankel matrix of the form $$ H_{n}:=\begin{pmatrix} h_{0} & h_{1} & \dots & h_{n-1}\\ h_{1} & h_{2} & \dots & h_{n}\\ \vdots & \vdots & \ddots & \vdots\\ h_{n-1} & h_{n} & \dots & h_{2n-2} \end{pmatrix}. $$ Just by a mathematical curiosity, I am interested in formulas for $\det H_{n}$ where $$ h_{n}=\binom{pn}{n} $$ where $p\geq2$ is an integer.

If $p=2$, the Hankel determinant (also called the Hankel transform) of central binomial coefficients is well-known and is not very difficult to evaluate. It reads $$ \det H_{n}=2^{n-1}. $$

For $p=3$, the formula (and its derivation) is more complicated, though it is also known that $$ \det H_{n}=3^{n-1}\left(\prod_{i=0}^{n-1}\frac{(3i+1)(6i)!(2i)!}{(4i)!(4i+1)!}\right), $$ see, for example, this paper for the proof.

Nevertheless, if $p\geq4$, I have not found any result about the corresponding Hankel determinant, nor I was able to even guess a form of the possible identity for $p=4$ while experimenting with Mathematica.

My questions are as follows:

  1. Is there something known about $\det H_{n}$ for $p\geq4$?
  2. Is anybody able to at least guess a formula for the case $p=4$?
  3. Is there any reason which might indicate that an explicit formula for $\det H_{n}$, with $p\geq4$, (like the one for $p=3$) need not exist?

Remark: I would like to remark that I know Krattenthaler's survey on advanced determinant calculus but I didn't find an answer therein.

  • 3
    $\begingroup$ One thing you can say about it: since ${4n \choose n}$ is divisible by $4$ for $n \ge 1$, the $2$-adic order $\nu_2(\det H_n) \ge 2n-2$. That inequality is an equality for $n = 1, 2, 4, 5, 6, 16, 17, 18, 20, 21, 22, 64, 65, 66, 68, 69, 70, 80, 81, 82, 84, 85, 86, \ldots$. Perhaps for all powers of $4$? $\endgroup$ Dec 1, 2016 at 19:37
  • 2
    $\begingroup$ Is $H_n > 0$ for all $n$ regardless of the value of $p$? This is, I think, equivalent to asking whether ${pn \choose n}$ is always a moment sequence, and in that case there is a standard technique to try related to orthogonal polynomials (see e.g. qchu.wordpress.com/2012/09/18/…). When $p = 2$ we get the moment sequence of an arcsin distribution; I already have no idea what distribution we get when $p = 3$, although the Hankel determinants do appear to be positive in this case. $\endgroup$ Dec 2, 2016 at 4:37
  • 1
    $\begingroup$ Perhaps it would be easier to consider instead of $\binom{4n}{n}$ more generally $\sum\binom{2n}{j}\binom{2n}{n-j}t^j.$ $\endgroup$ Dec 2, 2016 at 11:53
  • 4
    $\begingroup$ Note that ${np \choose n}$ is a moment sequence. It follows from the "sin-omials" solution that ${np \choose n}=\mathbb{E}(Y_{1/p})^{np}=\mathbb{E}(Y_{1/p}^p)^n$ where $Y_\alpha=\frac{\sin(X)}{(\sin(\alpha X))^\alpha (\sin((1-\alpha)X))^{1-\alpha}}$ and $X$ is uniform on $[0,\pi]$. Thus $\det(H_n)>0$ (the distribution of $Y_{1/p}$ is not finitely supported). $\endgroup$
    – esg
    Dec 7, 2016 at 19:57
  • 3
    $\begingroup$ One can also ask for the Smith normal form (SNF) when $p=2$ and $p=3$. For $p=2$ this is easy. For $p=3$ it is an open problem. One intriguing observation when $p=3$ is that the number $d_n$ of 3's in the SNF satifies $|d_n-\lfloor \frac 23n\rfloor|\leq 1$ for $n<224$, but this fails for $n=224$. See page 5 of math.mit.edu/~rstan/papers/snf_survey.pdf. $\endgroup$ Feb 14, 2017 at 23:02

1 Answer 1


This is not an answer.

Although there might be no "closed formula" for the case $p=4$, I wish to add the following to the case $p=2$ for which the determinants evaluate to $2^{n-1}$.

Suppose we decide to compute the determinant of the $n\times n$ Hankel matrix with left-most corner is at any position; in detail, consider the matrix $H_n(a)$ whose $(i,j)$ entry is given by $$\binom{2(i+j+a)}{i+j+a} \qquad \text{for} \qquad 0\leq i,j\leq n-1$$ then we have $$\det H_n(a)=\prod_{j=0}^{n-1}\binom{2a+2j}{a+j}\binom{a+n+j-1}{2j}^{-1}.$$ In particular, the original matrix $H_n=H_n(0)$ and we get a peculiar identity $$\prod_{j=0}^{n-1}\binom{2j}j\binom{n+j-1}{2j}^{-1}=2^{n-1}.$$


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