Fortune’s conjecture

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

 $n\#=\prod_{i=1}^{\pi(n)}p_{i}$

(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 $n\#+1$) is always a prime number. That is, any Fortunate number is a Fortunate prime.

It is obvious that since $n\#$ is divisible by each prime $p, then each $n\#+p$ will also be divisible by that same $p$ and thus not prime. If there is a prime $q>n\#+1$ such that there is a composite number $m=q-n\#$, 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 $(\log n\#)^{2}$ long immediately following a primorial, something he considers unlikely.

References

• 1 S. W. Golomb, “The evidence for Fortune’s conjecture,” Math. Mag. 54 (1981): 209 - 210. MR 82i:10053
Title Fortune’s conjecture FortunesConjecture 2013-03-22 17:31:17 2013-03-22 17:31:17 PrimeFan (13766) PrimeFan (13766) 4 PrimeFan (13766) Conjecture msc 11A41