# Simplify a factorial

I have the problem to evaluate the following:

$$(2n)!\over 2^n(n!)$$

Does this reduce to anything in particular?

I stuck it into a computer and it's

1: 1
2: 3
3: 15
4: 105
5: 945
6: 10395


No pattern immediately apparent.

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Have you noticed that each term divides the next term? –  Qiaochu Yuan Oct 2 '12 at 2:32
Were you computing the possible arrangements of a commutative, non-associative, operator over n terms, by any chance? –  Philippe Oct 2 '12 at 9:34
@Philippe No, it's just an exercise from a book. –  Luigi Plinge Oct 2 '12 at 22:09
@LuigiPlinge OK.. My colleague's whiteboard has the same sequence written all over it, complete with lots of little hand-drawn trees, that's why I was asking :) Also, for such things the On-Line Encyclopedia of Integer Sequences is a great resource. –  Philippe Oct 3 '12 at 8:02

Since $(2n)! = (2n) \times (2n-1) \times \cdots \times 2 \times 1$. Split the product into products of even factors and odd factors: $$(2n)! = \prod_{m=1}^{n} (2m) \cdot \prod_{m=1}^{n} (2m-1) = 2^n \prod_{m=1}^n m \cdot \prod_{m=1}^{n} (2m-1) = 2^n n! \prod_{m=1}^{n} (2m-1)$$ Therefore: $$\frac{(2n)!}{2^n n!} = \prod_{m=1}^n (2m-1) = (2n-1)!!$$ where $m!!$ denotes double factorial.

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Great, the book doesn't mention double factorials but this must be what they were alluding to. Thanks! –  Luigi Plinge Oct 2 '12 at 3:02

You won’t get a nice closed form, but there is another way to write it that is sometimes useful. Notice that

\begin{align*}2^nn!&=\underbrace{2\cdot2\cdot2\cdot\ldots\cdot2}_n\cdot1\cdot2\cdot3\cdot\ldots\cdot n\\&=(2\cdot1)(2\cdot2)(2\cdot3)\dots(2\cdot n)\\&=2\cdot4\cdot6\cdot\ldots\cdot 2n\;,\end{align*}

do some cancelling, and look at Qiaochu’s comment.

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You can use the identity

$$(2n)! = \Gamma(2n+1) = {\frac {{2}^{2n} \Gamma \left( n + 1\right) \Gamma \left( n + \frac{1}{2} \right) }{\sqrt {\pi }}}\,.$$

$$\frac{(2n)!}{2^n n!} = \frac{ 2^n \Gamma(n+\frac{1}{2})}{\sqrt{\pi}} = \frac{ 2^n (n-\frac{1}{2})!}{\sqrt{\pi}}\,.$$