Given some variables $x\_1, x\_2, \dots, x\_n$, the Vandermonde determinant is given by
$$V\_n(x\_1,\dots,x\_n):=\det(x\_j^{i-1})\_{i,j=1}^n=\prod\_{i<j}(x\_j-x\_i).$$
One can take as special cases: $x\_j=j$ or $x\_j=q^j$.


I was playing around with the following variation $A\_n=((i+j-1)^{i-1})\_{i,j=1}^n$. It turns out that (numerically) $\det(A\_n)=\det(j^{i-1})\_{i,j=1}^n$.


In addition, if $B\_n=(q^{(i+j-1)(i-1)})\_{i,j=1}^n$ then $\det(B\_n)$ and $\det(q^{j(i-1)})\_{i,j=1}^n$ differ only by a factor of a power of $q$, that is, $q^c$ for some $c$.


On the other hand, if $C\_n=(x\_{i+j-1}^{i-1})\_{i,j=1}^n$ then (unfortunately) $\det(C\_n)$ and $V\_n$ are vastly different; in fact, $\det(C\_n)$ does not factor as $V\_n$.



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> **QUESTION 1.** Is this true? $\det(A\_n)=\det(j^{i-1})\_{i,j=1}^n$.
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> **QUESTION 2.** What worked right for $A\_n$ andd $B\_n$ and failed for $C\_n$?
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A variant of numeric Vandermonde which failed symbolically?
Matrix $A\_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.


Since
$$(j+f(i))^{i-1} = \sum\_{k=0}^{i-1} c\_{i,k}\cdot j^k,$$
where $c\_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have
$$\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n = \begin{bmatrix} 
c\_{1,0} & 0 & 0 & \dots & 0\\
c\_{2,0} & c\_{2,1} & 0 & \dots & 0\\
c\_{3,0} & c\_{3,1} & c\_{3,2} & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
c\_{n,0} & c\_{n,1} & c\_{n,2} & \dots & c\_{n,n-1}
\end{bmatrix}\cdot \big[j^{i-1}\big]\_{i,j=1}^n.$$
Taking determinants, we get
$$\det\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n = c\_{1,0}c\_{2,1}\cdots c\_{n,n-1}\cdot \det\big[j^{i-1}\big]\_{i,j=1}^n=\det\big[j^{i-1}\big]\_{i,j=1}^n.$$




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For $B\_n$ things work even for a simpler reason:
$$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$
and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B\_n$.


However for matrix $C\_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite visual similarity $C\_n$ does not possess the properties of $A\_n$ or $B\_n$.


