Timeline for Logconcavity of height of Dyck paths
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 16, 2020 at 12:00 | history | bounty ended | CommunityBot | ||
| S Jun 16, 2020 at 12:00 | history | notice removed | CommunityBot | ||
| Jun 13, 2020 at 20:21 | comment | added | Brian Hopkins | I have an idea for a combinatorial approach that I'll just sketch for now. (1) Given two height $k$ paths, follow the first path until you reach height $k$, then insert the entire second path, and finally complete the first path. (2) Given a height $k-1$ path and a height $k+1$ path, build the same kind of composite path by interrupting the first at height $k-1$. Both of these approaches create height $2k$ paths. Why does (1) generate more? Or starting from the height $2k$ paths, why are there more ways to split them into two height $k$ paths than heights $k-1$ and $k+1$ paths? | |
| Jun 8, 2020 at 19:11 | answer | added | Per Alexandersson | timeline score: 2 | |
| S Jun 8, 2020 at 10:57 | history | bounty started | Mare | ||
| S Jun 8, 2020 at 10:57 | history | notice added | Mare | Draw attention | |
| Mar 26, 2020 at 20:55 | comment | added | Martin Rubey | A starting point is the product formula $(2k^2+6k+1-3n)(2n)!/((n-k)!(n+k+3)!)$ given in the reference, which is, however, valid only for $(n+1)/2\leq k\leq n$. Computing $T(n,k-1)T(n,k+1)/T(n,k)^2$ gives you three natural factors, which are all less than $1$. | |
| Mar 26, 2020 at 16:15 | history | asked | Mare | CC BY-SA 4.0 |