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 $a_n$


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?

  • $\begingroup$ What are the odd-indexed terms supposed to be? $\endgroup$ – Chris Eagle Mar 28 '12 at 18:28
  • 2
    $\begingroup$ Double factorial might be what you are looking for. $\endgroup$ – Dejan Govc Mar 28 '12 at 18:32
  • $\begingroup$ Well... that is indeed what I was looking for. Didn't know that notation existed! $\endgroup$ – cemulate Mar 28 '12 at 18:36
  • $\begingroup$ The next time you encounter a sequence of integers, it would benefit you greatly to look them up first in the OEIS; in particular, searching for the sequence $1, 2, 3, 8, 15, 48, 105, 384, 945, 3840,\dots$ gives this. $\endgroup$ – J. M. is a poor mathematician Mar 28 '12 at 19:26

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.

  • 1
    $\begingroup$ ...and analogously there are "triple factorials", "quadruple factorials", and so on. On that note: there are the convenient relations $(2n)!!=2^n n!$ and $(2n)!!(2n+1)!!=(2n+1)!$. $\endgroup$ – J. M. is a poor mathematician Mar 28 '12 at 18:47
  • $\begingroup$ @J.M. I was about to add all of that. But, my bad, the cases environment looks horrible. I'll edit to add all of that. Thanks for reading. $\endgroup$ – user21436 Mar 28 '12 at 18:49

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.


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


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