# 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 ${(\mathrm{log}n\mathrm{\#})}^{2}$ 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 |
---|---|

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 |