Timeline for Factorization of permutations into two factors with fixed number of cycles, plus a placement condition
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Feb 2, 2016 at 14:43 | history | bounty ended | CommunityBot | ||
| S Feb 2, 2016 at 14:43 | history | notice removed | CommunityBot | ||
| S Jan 25, 2016 at 13:00 | history | bounty started | Marcel | ||
| S Jan 25, 2016 at 13:00 | history | notice added | Marcel | Draw attention | |
| Jan 21, 2016 at 11:34 | history | edited | Marcel | CC BY-SA 3.0 |
Changed title, made one condition more clear
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| Jan 18, 2016 at 22:17 | comment | added | Marcel | Actually each tau always has exactly $n$ cycles. I don't really think that makes it easy to count, though. | |
| Jan 18, 2016 at 19:59 | comment | added | Gerhard Paseman | Not being a combinatorialist, consider the following as a crazy idea with a high probability of failure: wlog n less than m, in which case the first tau "pushes up" the integers 1 through n into different cycles, and the second " pushes them down" again, meaning each tau has at least n cycles. This should make it easy to count. For some reason the names zigzag permutation and juggling jump to mind. If Richard Stanley doesn't have a suggestion for you, you might try looking up zigzag and juggling in the combinatorics literature. Gerhard "Could Be Wrong, Of Course" Paseman, 2016.01.18. | |
| Jan 18, 2016 at 16:30 | history | edited | Marcel | CC BY-SA 3.0 |
added 8 characters in body
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| Jan 18, 2016 at 12:44 | history | asked | Marcel | CC BY-SA 3.0 |