Superfactorial

Template:Short description Template:Use list-defined references 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.

Definition

The $n$th superfactorial ${s}{f}(n)$ may be defined as:Template:R/superscript $$\begin{array}{rl}{s}{f}(n)& =1!\cdot 2!\cdot \cdots n!=\prod _{i=1}^{n}i!=n!\cdot {s}{f}(n-1)\\ & =1^n\cdot 2^{n-1}\cdot \cdots n=\prod _{i=1}^n{i}^{n+1-i}\\ & =\frac{(n!{)}^{n+1}}{\prod _{i=1}^n{i}^i}=\frac{(n!{)}^{n+1}}{H(n)}\end{array}$$where $H$ is the hyperfactorial.

Following the usual convention for the empty product, the superfactorial of 0 is 1. The sequence of superfactorials, beginning with ${s}{f}(0)=1$, is:Template:R/superscript Template:Bi

Properties

Just as the factorials can be continuously interpolated by the gamma function, the superfactorials can be continuously interpolated by the Barnes G-function as $sf(n)=G(n+2)$ for all nonnegative integers.Template:R/superscript

According to an analogue of Wilson's theorem on the behavior of factorials modulo prime numbers, when $p$ is an odd prime number $${s}{f}(p-1)\equiv (p-1)!!(\mathrm{mod}p),$$ where $!!$ is the notation for the double factorial.Template:R/superscript

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