Fortune’s conjecture

(Reo F. Fortune) For any integer $n>0$, the difference between the primorial

$$n\mathrm{\#}=\prod _{i=1}^{\pi (n)}{p}_i$$

(where $\pi (x)$ is the prime counting function and $p_i$ is the $i$th prime number) and the nearest prime number above (excluding the possible primorial prime $n\mathrm{\#}+1$) is always a prime number. That is, any Fortunate number is a Fortunate prime.

It is obvious that since $n\mathrm{\#}$ is divisible by each prime , then each $n\mathrm{\#}+p$ will also be divisible by that same $p$ and thus not prime. If there is a prime $q>n\mathrm{\#}+1$ such that there is a composite number $m=q-n\mathrm{\#}$, then $m$ would have to have at least two prime factors both of which would have to be divisible by primes greater than $p_{\pi (n)}$.

Despite verification for the first thousand primorials, this conjecture remains unproven as of 2007. Disproof could require finding a composite Fortunate number. Such a number would have to be odd, and indeed not divisible by the first thousand primes. Chris Caldwell, writing for the Prime Pages, argues that by the prime number theorem, finding a composite Fortunate number is tantamount to searching for a prime gap at least ${(\log n\mathrm{\#})}^2$ long immediately following a primorial, something he considers unlikely.

Title	Fortune’s conjecture
Canonical name	FortunesConjecture
Date of creation	2013-03-22 17:31:17
Last modified on	2013-03-22 17:31:17
Owner	PrimeFan (13766)
Last modified by	PrimeFan (13766)
Numerical id	4
Author	PrimeFan (13766)
Entry type	Conjecture
Classification	msc 11A41