Timeline for Primes with given Hamming weight
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 18, 2021 at 9:32 | vote | accept | Jamai-Con | ||
| Mar 18, 2021 at 9:32 | vote | accept | Jamai-Con | ||
| Mar 18, 2021 at 9:32 | |||||
| S Feb 21, 2021 at 18:06 | history | bounty ended | CommunityBot | ||
| S Feb 21, 2021 at 18:06 | history | notice removed | CommunityBot | ||
| Feb 13, 2021 at 17:29 | answer | added | Thomas Bloom | timeline score: 3 | |
| Feb 13, 2021 at 16:59 | history | edited | Jamai-Con | CC BY-SA 4.0 |
deleted 3 characters in body
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| S Feb 13, 2021 at 16:58 | history | bounty started | Jamai-Con | ||
| S Feb 13, 2021 at 16:58 | history | notice added | Jamai-Con | Improve details | |
| S Jun 9, 2019 at 0:02 | history | bounty ended | CommunityBot | ||
| S Jun 9, 2019 at 0:02 | history | notice removed | CommunityBot | ||
| Jun 7, 2019 at 15:32 | history | edited | Jamai-Con |
edited tags
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| S May 31, 2019 at 22:22 | history | bounty started | Jamai-Con | ||
| S May 31, 2019 at 22:22 | history | notice added | Jamai-Con | Draw attention | |
| May 30, 2019 at 0:01 | comment | added | Jamai-Con | @alpoge: Is there a chance that you turn your comments into a full-fledged answer later on? | |
| May 29, 2019 at 23:00 | comment | added | François Brunault | @alpoge I think you're right: we need the fact that the statement is uniform in $k$. | |
| May 29, 2019 at 22:53 | comment | added | alpoge | [If you were to fix k and take x to infinity the exponential term would die way faster than the log term in the parentheses, unless I’ve misread something, which is why I said careful in the previous comment.] | |
| May 29, 2019 at 22:52 | comment | added | alpoge | Careful! Take x = q^{k/\mu_q} and apply Theorem 1.1. The term in parentheses is 1 + O((\log{x})^{-1/2+\eps}). For k >> 1 this will be > 1/2. Also for k >> 1 the thing it’s multiplied by will be > 10. So the left-hand side is > 1. You see that needing k sufficiently large comes from the fact that the exponential in the parentheses has to beat the error term (and in particular the implicit constant). Hope that makes sense! | |
| May 29, 2019 at 22:17 | comment | added | François Brunault | @Wojowu For which values of $n$ does this expression tend to infinity? We cannot expect to take $n=2$ for example since this would give infinitely many Fermat primes. | |
| May 29, 2019 at 21:55 | comment | added | Wojowu | The statement follows immediately from Theorem 1.1 in the linked paper if you take $q=2$ and $k=n$. You just have to note that the expression in the bracket tends to infinity as $x\to\infty$. | |
| May 29, 2019 at 21:40 | history | asked | Jamai-Con | CC BY-SA 4.0 |