Notation for factorial-type pattern with a skip/step of two instead of one?

Question

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?

Answer 1

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.

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

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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.

Answer 2

It is written $n!!$ and denoted the double factorial.

Answer 3

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.