Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p\#=2\cdot3\cdot5\cdots p$ denote the primorial and $N(x)$ the smallest prime greater than or equal to $x$. Then Fortune's conjecture is that $N(p\#+2)-p\#$ is prime for all $p$. (Heuristic: to be composite the difference must be greater than $p^2$ which is large compared to the average gap of size $p$.) This seems out of reach at the moment.

A somewhat weaker version asks if the difference is composite only finitely often. This, too, seems unassailable at present.

A much weaker version asks if $N(p\#+2)-p\#$ is prime infinitely often. Can this be proved with current technology?

Edit: to clarify, my question is about the third problem I mention: can it be solved, has it been solved, and if so what is the solution? It's 'obvious' that the second and third conjectures are correct (and the first one seems highly likely) but I'm interested in what's known.

share|improve this question
1  
Are you asking if a currently open problem can be proven with the current state of mathematics? –  Aryabhata Jul 5 '11 at 15:57
2  
@Aryabhata: No, I'm asking if a problem related to an open problem is itself open or not. The first problem is open; the second is presumably open; the third I do not know. –  Charles Jul 5 '11 at 16:12
1  
Can you please edit the question to clarify? Also, I am not sure if the open-problem tag is appropriate in that case... –  Aryabhata Jul 5 '11 at 16:16
2  
@Aryabhata: I really don't see what's unclear, or how I could change it to be more clear. At no point do I ask for the solution to Fortune's conjecture. Feel free to edit the post if you have an idea of how to make it clearer. –  Charles Jul 5 '11 at 16:27
2  
Can't help, but there is some data and some references at oeis.org/A005235 that might help someone who wants to take this one on. –  Gerry Myerson Jul 6 '11 at 5:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.