A386860 The total number of big descents in all parking functions of length n.
0, 0, 4, 75, 1296, 24010, 491520, 11160261, 280000000, 7716919716, 232190115840, 7582217051695, 267271301197824, 10120214355468750, 409827566090715136, 17679671788737097545, 809596873977295011840, 39228032245196478804616, 2005401600000000000000000, 107880615499838355594014931
Offset: 1
Examples
a(2) = 0 because in the 3 parking functions of length 2 (11, 12, 21), there are 0 descents where the difference is strictly greater than one. a(3) = 4 as of the 16 parking functions of length 3 (111, 112, 122, 121, 212, 221, 211, 123, 132, 213, 312, 231, 321, 113, 131, 311) the parking functions (131, 311, 312, 231) all each have one big descent. Thus the total number of big descents in all parking functions of length 3 is 4.
Links
- Amanda Priestley, Table of n, a(n) for n = 1..100
- Kyle Celano, Jennifer Elder, Kimberly P. Hadaway, Pamela E. Harris, Amanda Priestley, and Gabe Udell, Inversions in parking functions, arXiv:2508.11587 [math.CO], 2025.
Programs
-
Mathematica
A386860[n_] := Binomial[n-1, 2]*(n+1)^(n-2); Array[A386860, 20] (* Paolo Xausa, Aug 07 2025 *)
Formula
a(n) = binomial(n-1,2)*(n+1)^(n-2).
a(n) = A386861(n)*2/n. - Paolo Xausa, Aug 07 2025
Comments