<s>[INST] Double Factorial expression in terms of regular factorials 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?
 [/INST] 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!
</s>