Determinant of sparse Hilbert matrix

Question

It is known that the determinant of the Hilbert matrix of dimension $N$ with elements $$ H^N_{tr}=\frac{1}{t+r-1}, \quad t,r=1,\dots,N $$ namely of the form \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \dots \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \dots \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \dots \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \dots \\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} decreases exponentially to zero as $$ \det H^N \approx a_N N^{-1/4} (2\pi)^N 4^{-N^2} . $$ I was wondering how the scaling of this matrix would change if we considered instead its elements being of the form $$ H^{N,S}_{tr}=\frac{1}{t+r-1}, \quad t,r \in \{ 1, S+1, 2S+1, \dots, NS+1 \} $$ namely ``skipping'' some values, obtaining for instance, by taking $S=3$, a matrix of the form \begin{pmatrix} 1 & \frac{1}{4} & \frac{1}{7} & \frac{1}{10} & \dots \\ \frac{1}{4} & \frac{1}{7} & \frac{1}{10} & \frac{1}{13} & \dots \\ \frac{1}{7} & \frac{1}{10} & \frac{1}{13} & \frac{1}{16} & \dots \\ \frac{1}{10} & \frac{1}{13} & \frac{1}{16} & \frac{1}{19} & \dots \\ \frac{1}{13} & \frac{1}{16} & \frac{1}{19} & \frac{1}{22} & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}