Timeline for Combinatorial interpretation of Sylvester–Lipschitz formula?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2023 at 13:22 | comment | added | Claude Chaunier | I forgot to mention I also removed the $(-1)^{k-1}$ factor. | |
| Oct 23, 2023 at 12:01 | comment | added | Claude Chaunier | For $a=3$ and $1\le k\le10$ and further dividing by 6 we get $1, 3, 39, 1107, 54351, 4085883, 435847959, 62594829027, 11644113200031, 2723549731505163$. Those appear in oeis.org/A080635 at odd positions, i.e. as the "number of permutations on $2k-1$ letters without double falls and without initial falls". | |
| Oct 20, 2023 at 9:30 | comment | added | Peter Taylor | @SamHopkins, if $x$ and $x+1$ are not adjacent, they can be swapped to get another alternating permutation. $x$ can't be adjacent to both $x-1$ and $x+1$ because then you have two consecutive ascents or two consecutive descents. There's a bit of work to be done to show that double-counting doesn't occur, but intuitively this feels like the explanation. | |
| Oct 20, 2023 at 1:54 | comment | added | Sam Hopkins | Is there an easy combinatorial explanation why $A_{2k-1}$, the number of alternating permutations of length $2k-1$, is divisible by $2^{k-1}$? | |
| Oct 19, 2023 at 17:21 | history | asked | Timothy Chow | CC BY-SA 4.0 |