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I know that the sequence of $n!$ is $$n(n-1)(n-2)\cdots(2)(1)$$ but what would be the sequence of $n!!$?

(In the interest of clarity, this is also known as the double factorial, not to be confused with $(n!)!$, i.e. the factorial of $n!$.)

Would it be $n^2(n-1)^2(n-2)^2\cdots(2)^2(1)^2$?

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    $\begingroup$ Are you asking about $(n!)!$ or this? $\endgroup$ – J.G. Dec 15 '18 at 18:11
  • $\begingroup$ double factorial thanks for the link $\endgroup$ – laxattack Dec 15 '18 at 18:14
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Assuming you refering to the notation $n!!$ instead of $(n!)!$ this is the so-called double factorial defined as the following

$$n!!=\begin{cases}n\cdot(n-2)\cdots3\cdot1&,\text{for odd } n>0\\n\cdot(n-2)\cdots4\cdot2&,\text{for even } n>0\\1 &,\text{for } n=0,-1\end{cases}$$

This definition can be either rewritten as a product formulation similiar to the one of the factorial or it can be expressed in terms of the factorial. To be precise the double factorial $(2n-1)!!$ is given by

$$(2n-1)!!=\frac{(2n)!}{2^n n!}$$

which can be proved via inductiion. Another interesting expression is given by the Pochhammer Symbol connecting the double factorial with the Gamma Function. Thus it can be written as

$$(2n-1)!!=2^n\frac{\Gamma(n+1/2)}{\Gamma(1/2)}$$

Hopefully this gives you a little insight on what the double factorial is and especially how it is related to the factorial.

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for even n $n!!=\prod_{k=1}^{\frac{n}{2}} 2k=n(n-2)(n-4)...4.2$

for odd n $n!!=\prod_{k=1}^{\frac{n+1}{2}} 2k-1=n(n-2)(n-4)...3.1$

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