Timeline for What is the status on this conjecture on arithmetic progressions of primes?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2019 at 0:13 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added 4 characters in body; edited tags
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| Aug 11, 2018 at 17:55 | comment | added | Thomas Browning | The conjecture holds up to $p=19$. See sequence A088430 in the OEIS. | |
| Jan 31, 2017 at 13:24 | vote | accept | Gorka | ||
| Jan 31, 2017 at 8:20 | answer | added | Greg Martin | timeline score: 28 | |
| Jan 31, 2017 at 5:37 | comment | added | Włodzimierz Holsztyński | I am not sure about the complete history of the problem $\ P(p)=p?\ $ I know that Siemion Fajtlowicz proposed this conjecture in 1991/2 or earlier. At that time I've got an algorithm and coded a program which gave me $\ P(13)=13.\ $ Once again, I am not a specialist, I don't know the full history here. My feeling was that $\ P(17) < 17\ ($ perhaps $\ \le 15).\ $ I feel strongly that $\ P(p) < p\ $ for every prime $\ p>13;\ $ I'd even conjecture that $\ p-P(p)\rightarrow \infty\ $ for $\ p\rightarrow\infty$. | |
| Jan 31, 2017 at 0:24 | comment | added | Pat Devlin | A good question. If true, the proof would have to be very delicate (not at all like the regularity-lemma-like proofs of Green-Tao). If false, I think a proof would be extremely strange. Perhaps current techniques can give some lower bound on $P(p)$, but this too sounds tricky to me. See the following (especially the last page or so) for some discussion of other related results. people.maths.ox.ac.uk/~conlond/green-tao-expo.pdf | |
| S Jan 31, 2017 at 0:07 | history | suggested | Pat Devlin | CC BY-SA 3.0 |
Reworded parts for slight clarity and added the reference to the Green-Tao theorem.
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| Jan 31, 2017 at 0:01 | review | Suggested edits | |||
| S Jan 31, 2017 at 0:07 | |||||
| Jan 29, 2017 at 13:47 | comment | added | Yaakov Baruch | Regarding $p=11$: the smallest sequence is $11+n\times 210\times 7315048$ for $0\le n \le 10$. | |
| Jan 29, 2017 at 12:08 | comment | added | Sylvain JULIEN | Sure, I just wanted to point out it can't be greater. | |
| Jan 29, 2017 at 11:06 | comment | added | Yaakov Baruch | @SylvainJULIEN in fact since the sequence in the question starts at $p$ itself, its length can be $= p$ (the delta then has to be divisible by all primes $< p$). | |
| Jan 29, 2017 at 10:01 | comment | added | Sylvain JULIEN | I may say something silly, but the reason of such an arithmetic progression must be even and coprime with p. Suppose the length of this progression is a prime q>p, then it contains 2 multiples of p, and thus can't consist only of primes. | |
| Jan 29, 2017 at 9:40 | comment | added | Yaakov Baruch | Already $p=11$ seems pretty tough... | |
| Jan 28, 2017 at 23:12 | comment | added | GH from MO | I am not aware of anything in that direction. Even $P(p)\geq 3$ sounds difficult to me. | |
| Jan 28, 2017 at 23:07 | comment | added | Gorka | I guess, it sounds strong enough so that it would be famous if proven. Maybe some weaker results about $P(p)$ are known? | |
| Jan 28, 2017 at 23:05 | comment | added | GH from MO | Interesting conjecture. I am sure it is open. | |
| Jan 28, 2017 at 21:32 | history | edited | Gorka | CC BY-SA 3.0 |
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| Jan 28, 2017 at 21:24 | history | asked | Gorka | CC BY-SA 3.0 |