Timeline for A conjecture regarding prime numbers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 28, 2017 at 0:12 | comment | added | Pasten | @BasantaR.Pahari I think your formulation is very elegant and it makes the problem look more natural. | |
| Nov 27, 2017 at 21:27 | comment | added | BR Pahari | @Pasten Thanks for pointing out that the conjecture is well-known . My formulation looks different but no doubt is equivalent to Dressler's conjecture. I think the way I wrote the problem in terms of sets made it difficult to find references or literature already published pertaining to the problem. | |
| Nov 27, 2017 at 21:01 | vote | accept | BR Pahari | ||
| Nov 27, 2017 at 18:14 | history | edited | Lucia | CC BY-SA 3.0 |
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| Nov 27, 2017 at 18:10 | comment | added | Lucia | @Pasten: Ah, good to know that the conjecture is already in the literature. The observation I made above (on abc + gaps between primes) seems to have been recorded by Cochrane and Dressler (in Math. Comp.). | |
| Nov 27, 2017 at 18:00 | comment | added | Pasten | cf. Dressler's conjecture. Often misquoted as a consequence of abc, while in fact it is a well-known consequence of abc and existence of primes in short intervals. | |
| Nov 27, 2017 at 14:40 | comment | added | Lucia | @YaakovBaruch: Nice example! | |
| Nov 27, 2017 at 13:18 | comment | added | Yaakov Baruch | There is a simple sequence where $n−m<(\frac{3}{4}n)^{1/2}$: $n=\frac{3}{4}9^k(9^k−1)$ and $m=\frac{3}{4}(9^k−1)^2$. | |
| Nov 26, 2017 at 23:24 | comment | added | BR Pahari | Nice! I am surprised that it is beyond the reach of RH. But that explains why my elementary approach (since I don't have any analytic number theory background ) weren't getting me anywhere. Regardless, it's really satisfying to see that the claim is actually true assuming the conjectures you mentioned. Great work! | |
| Nov 26, 2017 at 18:09 | comment | added | GH from MO | I agree. In the second case either $p\leq m$ in which case we are done, or by Bertrand's postulate we get easily that $n=p$ is prime, and then taking any prime divisor $q\mid m$ leads us back to the first case. Nice (conditional) proof! | |
| Nov 26, 2017 at 17:58 | comment | added | Lucia | If $n$ is divisible by a prime larger then $m$, then won't $n$ have to be that prime? (or one would have $n\ge 2m$ and then it should be obvious) | |
| Nov 26, 2017 at 17:53 | comment | added | GH from MO | I think we need a tiny bit more than this. Assume that $m<n$, and they don't have the same radical. Then there is a prime $p$ such that either $p\mid m$ and $p\nmid n$, or $p\nmid m$ and $p\mid n$. In the first case it is automatic that $p\in P_n-P_m$. In the second case we need the extra condition that $p\leq m$ to conclude that $p\in P_m-P_n$. Am I missing something? | |
| Nov 26, 2017 at 17:42 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 26, 2017 at 17:38 | history | answered | Lucia | CC BY-SA 3.0 |