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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?

  • 0
    The next time you encounter a sequence of integers, it would benefit you greatly to look them up first in the [OEIS](http://oeis.org/); in particular, searching for the sequence $1, 2, 3, 8, 15, 48, 105, 384, 945, 3840,\dots$ gives [this](https://oeis.org/A006882).2012-03-28

3 Answers 3

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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.

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    @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.2012-03-28
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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.

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