Consider $f(x) = (x+c)!$ and $g(x) = (ax+c)!$, where $a, x, c \geq 0$. Is there any way I can write $g(x)$ based on $f(x)$. I mean if I can write $g(x) = f(x)r(a, x)$, what would $r(a, x)$ be like?
In other words: $g(ax+c)! = (ax+c)\times(ax+c-1)\times(ax+c-2)\times\dots\times(x+c+1)\times(x+c)!$
What I am looking for is a straightforward expression for $r(a,x)$:
$r(a,x) = (ax+c)\times(ax+c-1)\times(ax+c-2)\times\dots\times(x+c+1)$.
So, you can think of it as:
$r(a,x) = (ax+c)! - (x+c)!$ where $a > 1$. But I need something for $r(a,x)$ without the use of factorial.