Consider the matrix $$A_n = \left( {a(i,j)} \right)_{i,j = 0}^{n - 1}$$ with $$a(i,j) = \binom{(i-j+r)x+s}{n-1}$$ where $r,s$ are arbitrary numbers. Is there a simple proof of the following?
$$\det (A_n) = x^{2 \binom{n}{2}}$$
Edit: I don’t know what you consider as relevant. I was only curious if the following determinants are a special case of a more general fact:
$$ \begin{vmatrix} 2&1\\3&2\end{vmatrix}=1^2,\quad \begin{vmatrix} 3&1\\5&3\end{vmatrix}=2^2,\quad \begin{vmatrix} 4&1\\7&4\end{vmatrix}=3^2,\dots, \\ \begin{vmatrix} 6&3&1\\10&6&3\\15&10&6\end{vmatrix}=1^6,\quad \begin{vmatrix} 15&6&1\\28&15&6\\45&28&15\end{vmatrix}=2^6,\quad \begin{vmatrix} 28&10&1\\55&28&10\\91&55&28\end{vmatrix}=3^6,\dots. $$
Computer experiments said yes, but I could not find a simple proof.
Darij Grinberg’s answer led to the following simple proof:
$\binom{x_i-y_j+s}{n-1}=\sum_{k=0}^{n-1}\binom{x_i}{k}\binom{s-y_j}{n-1-k}$
implies $M_n=(\binom{x_i-y_j+s}{n-1})_{0\leq i,j\leq n-1}=B_nC_n$
with $B_n=(\binom{x_i}{j})_{0\leq i,j\leq n-1}$ and $C_n=(\binom{s-y_j}{n-1-i})_{0\leq i,j\leq n-1}.$
Since $\det(\binom{x_i}{j})_{0\leq i,j\leq n-1}=\frac{\prod_{0\leq i<j\leq n-1}(x_j-x_i)}{ \prod_{0\leq i\leq n-1}(i!)}$ we finally get $\det(\binom{x_i-y_j+s}{n-1})_{0\leq i,j\leq n-1}=\frac{\prod_{0\leq i<j\leq n-1}(x_j-x_i) \prod_{0\leq i<j\leq n-1}(y_j-y_i)}{ \prod_{0\leq i\leq n-1}(i!)^2}.$
If we choose $x_i=xi+r$ and $y_j=xj$ we get $\det (A_n) = x^{2 \binom{n}{2}}.$