Timeline for least prime in a arithmetic progression
Current License: CC BY-SA 3.0
21 events
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| Apr 27, 2022 at 16:34 | comment | added | Will Jagy | @Bruno He has some visible expressions $\log^2 x,$ which I prefer to $( \log x)^2 $ in any case; it would then appear that the typo was a single left parenthesis in the wrong place, switched with the letter $q.$ Furthermore, as the Monthly likely has the same preference, they had no need for parentheses unless the intention was $q \log q$ inside them, as you show. | |
| Apr 27, 2022 at 16:22 | comment | added | Bruno | I asked Joseph Oesterlé. There is indeed a typo in the Monthly article, he only proved $p\le 70 (q\log q)^2$, that is the same as Bach-Sorenson (with a larger constant). You can find the announcement of the Oesterlé's results in Astérisque 61 (in French) here. | |
| Sep 11, 2015 at 6:02 | comment | added | joro | OK, I saw it... | |
| Sep 10, 2015 at 17:40 | comment | added | Will Jagy | @joro see mathoverflow.net/questions/217956/… | |
| Sep 10, 2015 at 16:58 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Sep 10, 2015 at 14:06 | comment | added | joro | I cited your claim $p \leq 70 q (\log q)^2$ and the dispute in comments here hurt my question: mathoverflow.net/questions/217919/… Consider editing to avoid confusion. | |
| S Sep 1, 2013 at 21:45 | history | suggested | Michael Albanese | CC BY-SA 3.0 |
Replaced \\, and \\; with \, so math would render properly.
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| Sep 1, 2013 at 21:43 | review | Suggested edits | |||
| S Sep 1, 2013 at 21:45 | |||||
| Mar 15, 2012 at 7:06 | comment | added | user709 | In fact, the Bach-Sorenson bound has been improved. See math.uiuc.edu/~xiannan/QuaResL1ver2.pdf | |
| Nov 23, 2011 at 1:54 | comment | added | Will Jagy | Eric, yes, there seems to be a consensus that the Oesterle result was incorrectly reported, and the Bach and Sorensen result I quote in comment is best. | |
| Nov 22, 2011 at 23:03 | comment | added | Eric Naslund | @Will Jagy: I believe that it should be $70q^2\log^2 q$ rather then $70q\log^2 q$. The former follows from GRH, whereas the later, an improvement to something of the form $q^{1+\delta}$ for any $\delta$, is only conjectured, and if proven would be a very strong result. | |
| Nov 14, 2011 at 21:51 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Nov 14, 2011 at 21:19 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Nov 14, 2011 at 21:05 | comment | added | Will Jagy | Sorry to hear about the Oesterle thing. My sense is that the OP just wants $q^2$ for convenience. Meanwhile, it appears you are talking about Bach and Sorensen, ams.org/journals/mcom/1996-65-216/S0025-5718-96-00763-6/… with ERH bound $$ p \leq (1 + o(1)) (\phi(q) \log q)^2 $$ | |
| Nov 14, 2011 at 20:24 | comment | added | so-called friend Don | Two quick comments. It's maybe interesting to note that for prime $q$, it's easy to establish the Monthly bound: Any prime divisor $p$ of $2^q-1$ works. Also, I'm skeptical of the bound attributed to Oesterle. I'd wager that the best bounds, on ERH, are those in ams.org/journals/mcom/1996-65-216/ Note that their upper bound is bigger than $q^2$, and so doesn't settle the OP's problem. The discussion in this paper also suggests the result attributed to Oesterle is based on a misunderstanding or a typo. | |
| Nov 14, 2011 at 7:54 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Nov 14, 2011 at 7:32 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Nov 14, 2011 at 5:25 | comment | added | Will Jagy | I do not know. The article makes special mention of results with remainder 1, as you ask, but they say nothing special about $q$ prime. They made some effort for completeness, so I would say your best result is Xylouris (2009). | |
| Nov 14, 2011 at 5:19 | comment | added | M.B | but $2^{\varphi(q)}$ is much bigger then $q^2$. So do you know if one can get a better bound then $2^{\varphi(q)}$ assuming $q$ is a prime? | |
| Nov 14, 2011 at 5:18 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Nov 14, 2011 at 5:11 | history | answered | Will Jagy | CC BY-SA 3.0 |