The falling factorial can be shown to be related to the gamma function according to: $(x)_n = \frac{\Gamma(x+1)}{\Gamma(x-n+1)}$.
Consider a generalized falling factorial of the form: $(x)_{n\lambda}=x(x-\lambda)(x-2\lambda)...(x-(n-1)\lambda)$, where $\lambda$ is not a constant that can be factored out.
Question: Is there any modified ("gamma") function that produce similar relationship as that between the ordinary falling factorial and gamma function? Please, notice that the problem doesn't reduce to: $(x)_{n\lambda}=\lambda^{n}(\frac{x}{\lambda})_n $
Letting
$$A_k(n) = \int_{0}^\infty \frac{1}{k} e^{-x/k} x^{n/k} \, dx$$
we see, by integration by parts, that
$$A_k(n) = n A_k(n-k)$$
and furthermore $A_k(0) = 1$.
Therefore, for $n$ a multiple of $k$, we see by induction that $$A_k(n) = n \cdot (n-k) \cdot (n-2k) \cdot \cdots \cdot 1$$
and so:
$$n \cdot (n-k) \cdot (n-2k) \cdot \cdots \cdot (n-(i-1)k) = A_k(n) / A_k(n-ik).$$
In fact this holds for $n$ not a multiple of $k$ as well.