Timeline for Counting function for prime pair with bounded gaps between them
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2015 at 13:09 | comment | added | Liron Yedidsion | Thank you very much. I think I get it. If I had a rather tight lower bound as well than the result in the link would have helped. | |
| Jul 29, 2015 at 13:07 | vote | accept | Liron Yedidsion | ||
| Jul 28, 2015 at 21:52 | comment | added | Stanley Yao Xiao | @LironYedidsion The answer given in your link is not what you're looking for I don't think. Zhang's proof can be strengthened (as can all such similar results) to obtain an asymptotic lower bound for the number of prime pairs that are say at most 70 million apart. This bound is of course at most $x$, since $x$ is the length of the interval. What you are looking for is a polynomial bound for the parameter $h(m)$ which guarantees that there are infinitely many $m$-tuples of primes which are $h(m)$ apart, which is not at all the same thing. | |
| Jul 28, 2015 at 20:09 | comment | added | Liron Yedidsion | This is what I feared. It makes a proof that I am working on a bit weaker as it is based on a conjecture. However, I still don't understand how does that comply with the answer in this link mathoverflow.net/questions/176875/… by GH from MO. Can you clarify this for me please? What am I missing? | |
| Jul 27, 2015 at 21:37 | comment | added | Gerhard Paseman | Right now there is no polynomial guarantee that such a bound exists. The Hardy-Littlewood conjectures concerning prime k-tuples and associated calculations can tell you how many such are expected to appear (something like $Cx/(\log x)^{2n}$ for an effectively computable C ). If you are willing to take those conjectures into account, you may get what you want. Gerhard "And Hopefully What You Need" Paseman, 2015.07.27 | |
| Jul 27, 2015 at 18:49 | comment | added | Liron Yedidsion | Dear Stanley, Thank you for your answer. it is important for me to find a bound that is not exponential in n. I stumbled upon this answer which I think might be exactly what I need. However, not being a mathematician, it would help me a lot to get an approval for this bound. mathoverflow.net/questions/176875/… | |
| Jul 27, 2015 at 16:43 | comment | added | Sylvain JULIEN | And maybe the OP would be glad to know that conjecturally, one can take $h(m)=m\log m$. | |
| Jul 27, 2015 at 13:05 | history | answered | Stanley Yao Xiao | CC BY-SA 3.0 |