Double Factorial

The double factorial is a generalization of the usual Factorial $n!$ defined by

$$n!!\equiv\cases{ n\cdot(n-2)\ldots 5\cdot 3\cdot 1 & $n$\ o... ...(n-2)\ldots 6\cdot 4\cdot 2 & $n$\ even\cr 1 & $n=-1, 0$.\cr}$$ (1)

For $n=0$, 1, 2, ..., the first few values are 1, 1, 2, 3, 8, 15, 48, 105, 384, ... (Sloane's A006882).

There are many identities relating double factorials to Factorials. Since

$(2n+1)!!2^nn!$
$= [(2n+1)(2n-1)\cdots 1][2n][2(n-1)][2(n-2)]\cdots 2(1)$
$= [(2n+1)(2n-1)\cdots 1][2n(2n-2)(2n-4)\cdots 2]$
$= (2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdots 2(1)$
$= (2n+1)!,$	(2)

it follows that $(2n+1)!! = {(2n+1)!\over 2^nn!}$. Since

$$(2n)!! = (2n)(2n-2)(2n-4)\cdots 2$$

$$= [2(n)][2(n-1)][2(n-2)]\cdots 2 = 2^nn!,$$ (3)

it follows that $(2n)!! = 2^nn!$. Since

$(2n-1)!!2^nn!$
$= [(2n-1)(2n-3)\cdots 1][2n][2(n-1)][2(n-2)]\cdots 2(1)$
$= (2n-1)(2n-3)\cdots 1][2n(2n-2)(2n-4)\cdots 2]$
$= 2n(2n-1)(2n-2)(2n-3)(2n-4)\cdots 2(1)$
$= (2n)!,$	(4)

it follows that

$$(2n-1)!! = {(2n)!\over 2^nn!}.$$ (5)

Similarly, for $n=0$, 1, ...,

$$(-2n-1)!! = {(-1)^n\over (2n-1)!!} = {(-1)^n2^nn!\over (2n)!}.$$ (6)

For $n$ Odd,

$${n!\over n!!} = {n(n-1)(n-2)\cdots (1)\over n(n-2)(n-4)\cdots (1)}$$

$$= (n-1)(n-3)\cdots (1) = (n-1)!!.$$ (7)

For $n$ Even,

$${n!\over n!!} = {n(n-1)(n-2)\cdots (2)\over n(n-2)(n-4)\cdots (2)}$$

$$= (n-1)(n-3)\cdots (2)= (n-1)!!.$$ (8)

Therefore, for any $n$,

$${n!\over n!!}=(n-1)!!$$ (9)

$$n!=n!!(n-1)!!.$$ (10)

The double factorial is a special case of the Multifactorial.

See also Factorial, Multifactorial

References

Sloane, N. J. A. Sequence A006882/M0876 in ``An On-Line Version of the Encyclopedia of Integer Sequences.'' http://www.research.att.com/~njas/sequences/eisonline.html and Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego: Academic Press, 1995.