Notation for factorial-type pattern with a skip/step of two instead of one? I came across a peculiar pattern when solving a recurrence relation today:
Some sequence $a_n$ looks as such:
$a_0 = 1$
$a_2 = \frac{1}{2 \cdot 1}$
$a_4 = \frac{1}{4 \cdot 2 \cdot 1}$
$a_6 = \frac{1}{6 \cdot 4 \cdot 2 \cdot 1}$
...
The pattern is quite simple, but I can't think of a way to express the general equation for $a_n$
EDIT:
The recurrence relation is
$a_{n+2} = \frac{1}{n+2} \cdot a_n$
so the odd indices' denominators would be 1, 3, 5*3, 7*5*3, etc.
Any pointers?
 A: You are looking for probably this piece of notation, called double factorial:
$$n!!= \begin{cases}n \cdot (n-2) \cdot (n-4) \cdots 3 \cdot 1, &\text{$n \gt 0$, n odd}\\ n \cdot (n-2) \cdot (n-4) \cdots 4 \cdot 2, &\text{$n \gt 0$, n even} \\ 1,  &\text{$n=-1,0$}\end{cases}$$
Most conveniently, we would write it as $$\begin{align}(2n)!!&=2 \cdot 4 \cdot 6 \cdots 2n~~ \mbox{for}~~ n \in \Bbb N \\(2n+1)!!&=1 \cdot 3 \cdot 5 \cdots (2n+1)~~ \mbox{for}~~ n \in \Bbb N\end{align}$$
Some relations connecting these quantities:


*

*$(2n)!!=\prod_{i=1}^n(2i)=2^n\prod_{i=1}^n i=2^nn!$

*$(2n+1)!!(2n)!!=(2n+1)!$ obviously.

For the problem at hand, your terms are $$a_n=\dfrac 1 {n!!}$$

Analogously, one could define functions like primorial, triple factorial, quadruple factorials and so on. For a more extended note on this kind of definitions, please follow this link.
A: As mentioned, a slightly non-standard use of double-factorial gives
$$
a_n=\frac{1}{n!!}
$$
for even $n$. However, I usually see double factorial used with odd $n$. For even $n$,
$$
a_n=\frac{1}{2^{n/2}(n/2)!}
$$
works as well.
For odd $n$, without double factorial,
$$
a_n=\frac{2^{(n-1)/2}(\frac{n-1}{2})!}{n!}
$$
so it is easy to see why $1/n!!$ is preferred.
A: It is written $n!!$ and denoted the double factorial.
