I have the following double factorial expression: $$\frac{(x-2)!!(x-2k-1)!!}{(x-1)!!(x-2k)!!}$$ if x is an odd number

I am trying to rewrite this without double factorials by using the following expressions for double factorials: $$n!!=2^{n/2}(n/2)!$$ for even $n$ and $$\frac{(n+1)!}{2^{(n+1)/2}(\frac{n+1}{2})!}$$ for odd $n$.

What is the equivalent expression using these definitions? Is there a better way to write the original expression without double factorials?

  • $\begingroup$ Simplify $\frac{n!!}{(n+1)!!}$ first. $\endgroup$
    – J.G.
    Apr 3, 2019 at 22:37
  • $\begingroup$ I get that for even n, $\frac{n!!}{(n+1)!!}=\frac{2^{n+1}*(n/2)!*(\frac{n+2}{2})!}{(n+2)!}$. For odd n, $\frac{n!!}{(n+1)!!}=\frac{n!}{2^{(n/2)!}*(n/2)!}$ $\endgroup$ Apr 3, 2019 at 23:16

1 Answer 1


First of all: $$ \frac{(x-2)!!(x-2k-1)!!}{(x-1)!!(x-2k)!!} = \frac{(x-2)!!}{(x-1)!!} \times \frac{(x-2k-1)!!}{(x-2k)!!} $$

For the first factor:

$$ \begin{align} \frac{(x-2)!!}{\left(x-1\right)!!}& = \frac {\frac{(x-1)!}{2^{(x-1)/2}(\frac{x-1}{2})!}} {2^{(x-1)/2}\left(\frac{x-1}2\right)!}\\ & = \frac{(x-1)!}{\left(2^{(x-1)/2}(\frac{x-1}{2})!\right)^2}\\ & = \frac{(x-1)!}{2^{x-1}\left(\left(\frac{x-1}{2}\right)!\right)^2} \end{align} $$

A similar sequence with the second factor gives:

$$ \begin{align} \frac{(x-2k-1)!!}{(x-2k)!!}& = \frac {2^{(x-2k-1)/2}\left(\frac{x-2k-1}2\right)!} {\frac{(x-2k+1)!}{2^{(x-2k+1)/2}\left(\frac{x-2k+1}{2}\right)!}}\\ & = \frac {2^{(x-2k-1)/2}\left(\frac{x-2k-1}2\right)! \; {2^{(x-2k+1)/2}\left(\frac{x-2k+1}{2}\right)!}} {(x-2k+1)!}\\ & = \frac {2^{x-2k}\left(\frac{x-2k-1}2\right)! \; {\left(\frac{x-2k+1}{2}\right)!}} {(x-2k+1)!} \end{align} $$

Multiply them together and get:

$$ \frac {2^{x-2k}\left(\frac{x-2k-1}2\right)! \; {\left(\frac{x-2k+1}{2}\right)!}} {(x-2k+1)!} \frac{(x-1)!}{2^{x-1}\left(\left(\frac{x-1}{2}\right)!\right)^2} = \frac{2^{x-2k}}{2^{x-1}} \frac {\left(\frac{x-2k-1}2\right)!} {\left(\frac{x-1}{2}\right)!} \frac{\left(\frac{x-2k+1}{2}\right)!}{\left(\frac{x-1}{2}\right)!} \frac{(x-1)!}{(x-2k+1)!} = 2^{1-2k} \frac {\left(\frac{x-2k-1}2\right)!} {\left(\frac{x-1}{2}\right)!} \frac{\left(\frac{x-2k+1}{2}\right)!}{\left(\frac{x-1}{2}\right)!} \frac{(x-1)!}{(x-2k+1)!} $$

That is the best I will do for now. The ratio of two factorials (with occurs three times in this expression) often has meaning as a product of consecutive integers, not beginning with one.

Hope this helps!


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