The double factorial is defined as $$n!! = \begin{cases} n \cdot (n-2) \cdot (n-4) \cdots 3 \cdot 1 = \dfrac{(n+1)!}{2^{(n+1)/2}((n+1)/2)!} & \text{ If $n \in \mathbb{Z}^+$, is odd}\\ n \cdot (n-2) \cdot (n-4) \cdots 4 \cdot 2 = 2^{n/2} (n/2)! & \text{ If $n \in \mathbb{Z}^+$, is even}\\ 1 & \text{If $n \in \{0,-1\}$} \end{cases} \,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ I am curious about the origin of this notation. I see it to be a poor, non-suggestive notation. After a quick search, I see that the notation was first used in Arfken $1985$. I am curious as to why one would use such a non-suggestive notation, call it double factorial and more importantly, why have others caught on to this notation? Also, are there alternate notations to denote $(\star)$?

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    $\begingroup$ Curious question. When you look at the references on MathWorld, they mention Meserve, B. E. "Double Factorials from 1948. Also, you can see it on OIES. Additionally, we get into Multi Factorial. $\endgroup$
    – Amzoti
    Jan 2, 2013 at 23:20
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    $\begingroup$ @AlexanderGruber The notation $n!_{k}$ seems to make sense and is at the least better than the multi-factorial notation. $\endgroup$
    – user17762
    Jan 2, 2013 at 23:27
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    $\begingroup$ I've always thought this notation is sub-optimal because it looks like an iterated factorial---a factorial of a factorial. $\endgroup$ Jan 3, 2013 at 0:42
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    $\begingroup$ I personally think the notation is fine (perhaps because I learned it too long ago). The operation is related to factorial and the operation is more commonly encountered than $(n!)!$. I think of it as the "every other" factorial. Interesting question. (+1) $\endgroup$
    – user26872
    Jan 3, 2013 at 2:25
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    $\begingroup$ Continuing along the lines of "every other" factorial, I first encountered the notation in the context of perfect matchings. A complete graph on $n+1$ vertices ($n+1$ even) has $n!!$ perfect matchings. Choosing an arbitrary vertex as "1", there are $n$ other vertices to pair it with. Now we have a complete graph on $n-1$ vertices, which has $(n-2)!!$ perfect matchings by induction. Therefore, the original graph has $n(n-2)!! = n!!$ perfect matchings. (Every choice we make knocks out two vertices, hence the "every other" aspect.) $\endgroup$ Jan 3, 2013 at 2:59

2 Answers 2


Well, the way I assume it was chosen was:

With one factorial sign, we decrement by one:

$n! = n \cdot (n - 1) \cdot ... \cdot 2 \cdot 1$

So, if we have two factorial signs, let's decrement by two!

$n!! = n \cdot (n - 2) \cdot ... \cdot 3 \cdot 1$

And thus, it was done.

It's a terrible abuse of notation, as mentioned in the other answer, I'll agree.


The way I parse it, $n!$ is a post-fix function call. Therefore $n!!$ parses as factorial(factorial(n)). This notation has my disapprobation.

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    $\begingroup$ I think this should be a comment. $\endgroup$ Jan 3, 2013 at 2:55
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    $\begingroup$ It solicits opinions on notation, so I see it as reasonable to post this as an answer. I am addressing the question directly, which is a matter of opinion. $\endgroup$ Jan 3, 2013 at 2:57
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    $\begingroup$ @ncmathsadist except that the question explicitly asks for "the origin of this notation" and "why one would use such a non-suggestive notation". $\endgroup$
    – Cole Tobin
    Apr 17, 2014 at 15:24

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