Superfactorial

In mathematics, and more specifically number theory, the superfactorial of a positive integer $n$ is the product of the first $n$ factorials. They are a special case of the Jordan–Pólya numbers, which are products of arbitrary collections of factorials.

The $n$ th superfactorial ${\mathit {sf}}(n)$ may be defined as:^[1] ${\begin{aligned}{\mathit {sf}}(n)&=1!\cdot 2!\cdot \cdots n!=\prod _{i=1}^{n}i!=n!\cdot {\mathit {sf}}(n-1)\\&=1^{n}\cdot 2^{n-1}\cdot \cdots n=\prod _{i=1}^{n}i^{n+1-i}.\\\end{aligned}}$ Following the usual convention for the empty product, the superfactorial of 0 is 1. The sequence of superfactorials, beginning with ${\mathit {sf}}(0)=1$ , is:^[1]

1, 1, 2, 12, 288, 34560, 24883200, 125411328000, 5056584744960000, ... (sequence A000178 in the OEIS)

Just as the factorials can be continuously interpolated by the gamma function, the superfactorials can be continuously interpolated by the Barnes G-function.^[2]

According to an analogue of Wilson's theorem on the behavior of factorials modulo prime numbers, when $p$ is an odd prime number ${\mathit {sf}}(p-1)\equiv (p-1)!!{\pmod {p}},$ where ${\displaystyle$ is the notation for the double factorial.^[3]

For every integer $k$ , the number ${\mathit {sf}}(4k)/(2k)!$ is a square number. This may be expressed as stating that, in the formula for ${\mathit {sf}}(4k)$ as a product of factorials, omitting one of the factorials (the middle one, $(2k)!$ ) results in a square product.^[4] Additionally, if any $n+1$ integers are given, the product of their pairwise differences is always a multiple of ${\mathit {sf}}(n)$ , and equals the superfactorial when the given numbers are consecutive.^[1]