What would be the sequence of $n!!$ (double factorial)

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 $$n^2(n-1)^2(n-2)^2\cdots(2)^2(1)^2$$?

• Are you asking about $(n!)!$ or this? – J.G. Dec 15 '18 at 18:11
• double factorial thanks for the link – 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

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

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

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

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