$2\times 5 \times 8 \ldots \times (3n-1)=?$

Question

Does anybody know if there is a closed form expression using factorials for the above product? I'm not seeing it but I feel like there must be. The recursive relationship corresponding to this expression is just $a_1=2$ and $a_n= (3n-1)a_{n-1}$ but like I said, I'm not seeing it. Any and all insights are welcome.

Answer 1

This looks very similar to Pochhammer symbol used to represent the falling factorial $$(x)_n=x(x-1)(x-2)\cdots(x-n+1)$$ and when $x$ is a non-negative integer, $(x)_n$ gives the number of $n$ permutations of an $x$ element set, or equivalently the number of injective functions from a set of size $n$ to a set of size $x$ (this is a quote from the corresponding Wikipedia page).

So $$\prod_{k=1}^n (a k-1)=a^n \left(\frac{a-1}{a}\right)_n=a^n\frac{ \Gamma \left(n+1-\frac{1}{a}\right)}{\Gamma \left(1-\frac{1}{a}\right)}$$ In the case of the post $a=3$.

For the recursion given in the post, the formal result is $$u_n=2\ 3^{n-1} \left(\frac{5}{3}\right)_{n-1}$$

Answer 2

The multifactorial notation I just found being used online is $$n!^{(b)}=\prod_{k=0}^{\lfloor{\frac{n}{b}\rfloor}} (n-bk)$$ Can someone confirm that this is standard?