Timeline for Counting faces on multipermutahedra/multipermutohedra
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| May 3, 2016 at 23:46 | comment | added | Richard Stanley | @VinceMatsko: you are right. I am counting all faces, while you are counting the different combinatorial types. | |
| Apr 3, 2016 at 21:23 | comment | added | Vince Matsko | Let me clarify with an example. Consider the $n$-dimensional simplex $0^n1.$ The number of ordered partitions into $k$ blocks should always be $1$ since there is only one type of face of a given dimension on a simplex. When I calculate the count directly using the formula above, I find that the coefficient of $x_0^nx_1$ in $((x_0+x_0^2+x_0^3+\cdots)(x_1+x_1^2+x_1^3+\cdots)+x_0+x_1)^k$ is $k$ for $k\le n+1$ rather than $1.$ When $n=4$ and $k=2,$ for example, I should only get $0|0001.$ I am not sure we are both counting the same thing. | |
| Apr 3, 2016 at 1:30 | history | answered | Richard Stanley | CC BY-SA 3.0 |