Timeline for Update for 2015: least prime of form nq+1, with q prime?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2022 at 16:28 | comment | added | Bruno | As mentioned below, there is a typo in reporting the "private communication" according to J. Oesterlé himself. He proved $p\le 70 (q\log q)^2$, not $p\le 70 q(\log q)^2$. Cf my comment mathoverflow.net/questions/80865/…. | |
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 11, 2015 at 8:13 | comment | added | joro | The $70$ appears in this paper in "private communication": hri.res.in/~thanga/papers/final-amm.pdf | |
| Sep 10, 2015 at 19:51 | answer | added | Igor Rivin | timeline score: 2 | |
| Sep 10, 2015 at 19:24 | history | edited | Lucia |
Added top level number theory tag
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| Sep 10, 2015 at 19:19 | vote | accept | Will Jagy | ||
| Sep 10, 2015 at 17:26 | answer | added | Lucia | timeline score: 17 | |
| Sep 10, 2015 at 17:22 | comment | added | Stanley Yao Xiao | To my knowledge the best known bound due to GRH is something like $p \leq q^{2 + \epsilon}$. If there is a Siegel zero, then an exponent strictly smaller than $2$ is possible. I believe the latter is due to Friedlander and Iwaniece. | |
| Sep 10, 2015 at 17:22 | history | edited | Yemon Choi | CC BY-SA 3.0 |
light edits to formatting and title
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| Sep 10, 2015 at 17:17 | history | edited | Will Jagy | CC BY-SA 3.0 |
added 132 characters in body
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| Sep 10, 2015 at 17:10 | history | asked | Will Jagy | CC BY-SA 3.0 |