TITLE: Hankel determinants of binomial coefficients
QUESTION [16 upvotes]: 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:

Is there something known about $\det H_{n}$ for $p\geq4$?
Is anybody able to at least guess a formula for the case $p=4$?
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.

REPLY [3 votes]: 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
$$\prod_{j=0}^{n-1}\binom{2j}j\binom{n+j-1}{2j}^{-1}=2^{n-1}.$$