Timeline for A Markov consensus
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2018 at 20:42 | comment | added | Hauke Reddmann | Some MATHEMATICA: g[n_, m_] := Count[Rest /@First /@Tally /@ IntegerPartitions[n] // Flatten, m]; will compute the number of tie multiplicities m in the list P(n), and f[n_, m_] := Select[Min /@ IntegerPartitions[n], # >= m &] // Length; will count all P(n) that have only elements >=m. As you can see in a printout, f==g. | |
| Jan 17, 2018 at 20:09 | comment | added | Hauke Reddmann | The general pattern seems to be (if I didn't mess up MATHEMATICA or OEIS): #m-tie in P[n]=#P[n-m]|only parts >=m. | |
| Jan 17, 2018 at 20:05 | comment | added | Arnaud Mortier | Beautiful! Let's see if we can build up on this to get a bound. | |
| Jan 17, 2018 at 20:02 | comment | added | Hauke Reddmann | @Arnaud: A quick check gave the following nice result: The number of partitions of n with untied winner are...the number of partitions of n-1. The number of partitions of n with exactly one tie are...the number of partitions of n-2 that do not contain 1 as a part. Triple tie - P(n-3) with parts>3. And so on? | |
| Jan 15, 2018 at 9:11 | history | edited | coudy | CC BY-SA 3.0 |
improved layout
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| Jan 15, 2018 at 0:03 | answer | added | Mateusz Kwaśnicki | timeline score: 5 | |
| S Jan 15, 2018 at 0:02 | history | suggested | Arnaud Mortier |
added tag co.c
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| Jan 14, 2018 at 23:33 | comment | added | Arnaud Mortier | A simpler question to get started would be how many of the maps $\left\lbrace 1\ldots n \right\rbrace \rightarrow \left\lbrace 1\ldots n \right\rbrace $ are so that one integer gets mapped to strictly more often than any other. I'd therefore add the tag combinatorics. | |
| Jan 14, 2018 at 23:27 | review | Suggested edits | |||
| S Jan 15, 2018 at 0:02 | |||||
| Jan 14, 2018 at 22:03 | history | asked | Hauke Reddmann | CC BY-SA 3.0 |