Timeline for A combinatorial proof for the property of KM numbers?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2020 at 19:58 | vote | accept | Mikhail Gaichenkov | ||
| S Apr 12, 2018 at 8:31 | history | bounty ended | Mikhail Gaichenkov | ||
| S Apr 12, 2018 at 8:31 | history | notice removed | Mikhail Gaichenkov | ||
| Apr 5, 2018 at 15:20 | answer | added | David E Speyer | timeline score: 8 | |
| S Apr 5, 2018 at 13:34 | history | bounty started | Mikhail Gaichenkov | ||
| S Apr 5, 2018 at 13:34 | history | notice added | Mikhail Gaichenkov | Draw attention | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Aug 24, 2013 at 21:18 | history | bounty ended | CommunityBot | ||
| S Aug 24, 2013 at 21:18 | history | notice removed | CommunityBot | ||
| S Aug 16, 2013 at 19:41 | history | bounty started | Mikhail Gaichenkov | ||
| S Aug 16, 2013 at 19:41 | history | notice added | Mikhail Gaichenkov | Draw attention | |
| Aug 30, 2011 at 8:15 | comment | added | Mikhail Gaichenkov | Well, let $I_n(k)$ - the number of permutations of $n$ objects with precisely $k$ inversions. We should study the property of the numbers $M(n)=I_n([n(n-1)/4])$. | |
| Aug 29, 2011 at 22:58 | comment | added | Brendan McKay | Do you agree that the description at OEIS, namely "Kendall-Mann numbers: the maximal number of inversions in a permutation on n letters is floor(n(n-1)/4); a(n) = number of permutations with this many inversions" is wrong? It should refer to the maximum number of permutations having the same number of inversions, right? | |
| Aug 29, 2011 at 12:12 | history | asked | Mikhail Gaichenkov | CC BY-SA 3.0 |