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

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


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


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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