This question has been inspired by Hankel determinants of binomial coefficients. For a sequence $\{h_{n}\}_{n=0}^{\infty}$ denote by $H_n$ the Hankel matrix

$$H_{n}:=\begin{pmatrix} h_{0} & h_{1} & \dots & h_{n}\\ h_{1} & h_{2} & \dots & h_{n+1}\\ \vdots & \vdots & \ddots & \vdots\\ h_{n} & h_{n+1} & \dots & h_{2n} \end{pmatrix}.$$

It is well known that the Hankel determinant $ \det{A_n}$ of the sequence $a_n=\binom{2n}{n}$ of central binomial coefficients satisfies $ \det A_{n}=2^{n}$ and the Hankel determinant $\det{B_n}$ of the sequence of Catalan numbers $b_n=\binom{2n}{n}\frac{1}{n+1}$ satisfies $ \det B_{n}=1.$ Thus $$\frac{\det{A_n}}{\det{B_n}}=2^n.$$ Computer experiments suggest that more generally for the sequences $a_n=\binom{kn}{n}$ and $b_n=\binom{kn}{n}\frac{1}{(k-1)n+1}$ for a positive integer $k$ we get $$\frac{\det{A_n}}{\det{B_n}}=k^n.$$ Is there a simple way to reduce $ \det{A_n}$ to $ \det{B_n}?$ It should be noted that for $k\ge4$ no explicit formulae are known.

**Edit**: In the mean-time I have found a proof using the J-fraction expansion of the generating functions. But it would be interesting if there is a direct proof using row and column operations.