Timeline for If $p_{n}$ is the largest prime factor of $p_{n-1}+p_{n-2}+m$, then $p_{n}$ is bounded
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Mar 28, 2017 at 5:13 | history | bounty ended | CommunityBot | ||
| S Mar 28, 2017 at 5:13 | history | notice removed | CommunityBot | ||
| Mar 22, 2017 at 22:41 | comment | added | Igor Rivin | I assume that the (alleged) result is the same if you take an arbitrary prime factor, not just the largest one... | |
| Mar 22, 2017 at 9:42 | answer | added | Elizabeth S. Q. Goodman | timeline score: 1 | |
| Mar 21, 2017 at 21:08 | comment | added | Igor Rivin | Where does this question come from? | |
| S Mar 20, 2017 at 3:41 | history | bounty started | math110 | ||
| S Mar 20, 2017 at 3:41 | history | notice added | math110 | Authoritative reference needed | |
| Mar 17, 2017 at 19:56 | answer | added | Robert Israel | timeline score: 5 | |
| Mar 17, 2017 at 16:00 | comment | added | Gerhard Paseman | Also it is not clear that $p_n + p_{n-1} +m$ cannot assume an unbounded number of prime values, although in this case it should be provable that it leads to large multiples of 3 also. Gerhard "Life Is Easier Modulo Three" Paseman, 2017.03.17. | |
| Mar 17, 2017 at 15:50 | comment | added | TonyK | @PéterKomjáth: I think that $p_n$ just means the $n^\mathrm{th}$ prime in the sequence, not the $n^\mathrm{th}$ largest prime. | |
| Mar 17, 2017 at 15:40 | comment | added | Péter Komjáth | As $p_{n-2},p_{n-1}$, and $p_n$ are approximately the same, the only possibility is that $p_{n-2}+p_{n-1}+m=2p_n$, so $m$ must be even. | |
| S Mar 17, 2017 at 10:56 | history | suggested | Glorfindel | CC BY-SA 3.0 |
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| Mar 17, 2017 at 10:25 | review | Suggested edits | |||
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| Mar 17, 2017 at 1:28 | review | First posts | |||
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| Mar 17, 2017 at 1:27 | history | asked | math110 | CC BY-SA 3.0 |