A333829 -id:A333829 - cp's OEIS frontend

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

Amanda Priestley, Aug 05 2025

A big descent in a parking function (x_1,x_2,...,x_k) is a position i such that x_i - x_{i+1} >= 2.

			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.

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.

Cf. A000272(n+1) (parking functions), A333829, A386015, A386861.

A386860[n_] := Binomial[n-1, 2]*(n+1)^(n-2);
Array[A386860, 20] (* Paolo Xausa, Aug 07 2025 *)

a(n) = binomial(n-1,2)*(n+1)^(n-2).

a(n) = A386861(n)*2/n. - Paolo Xausa, Aug 07 2025

A225753 Triangle of transformations with k monotonic runs.

Original entry on oeis.org

1, 3, 1, 10, 16, 1, 35, 155, 65, 1, 126, 1246, 1506, 246, 1, 462, 9142, 24017, 12117, 917, 1, 1716, 63792, 315918, 349840, 88852, 3424, 1, 6435, 432399, 3707559, 7635987, 4362297, 619677, 12861, 1, 24310, 2881450, 40455910, 140543458, 149803270, 49462810, 4200670, 48610, 1

Offset: 1

Chad Brewbaker, May 14 2013

Analogous to the Eulerian triangle for permutations A173018.
T(n,k) is the number of words of length n over the alphabet {0,1,...,n-1} that have k-1 descents, see example. [Joerg Arndt, Jun 25 2013]
The expected number of descents is (Sum_{k=1..n} (k-1)*T(n,k)) / (Sum_{k=1..n} T(n,k)) = (n + 1/n -2)/2. - Geoffrey Critzer, Jun 26 2013

			T(1,1) = #{[0]} = 1.
T(2,1) = #{[0,0], [0,1], [1,1]} = 3.
T(2,2) = #{[1,0]} = 1.
T(3,1) = #{[0,0,0], [0,0,1], [0,0,2], [0,1,1], [0,1,2], [0,2,2], [1,1,1], [1,1,2], [1,2,2], [2,2,2]} = 10.
Triangle T(n,k) begins:
     1;
     3,     1;
    10,    16,      1;
    35,   155,     65,      1;
   126,  1246,   1506,    246,     1;
   462,  9142,  24017,  12117,   917,    1;
  1716, 63792, 315918, 349840, 88852, 3424,  1;
  ...

Alois P. Heinz, Rows n = 1..100, flattened

First column is A001700(n-1).
Row sums give: A000312.
Cf. A008292, A062023, A190295, A226998, A333829.

b:= proc(n, l, k) option remember; local j;
      if n=0 then [1] else []; for j to k do zip((x, y)->x+y,
       %, [`if`(j b(n, 0, n)[]:
seq(T(n), n=1..10);  # Alois P. Heinz, Jun 26 2013

Table[Distribution[Map[Length,Map[Split[#,LessEqual[#1,#2]&]&,Tuples[Range[1,n],n]]]],{n,1,7}]//Grid (* Geoffrey Critzer, Jun 25 2013 *)
zip[f_, x_, y_, z_] := With[{m = Max[Length[x], Length[y]]}, f[PadRight[x, m, z], PadRight[y, m, z]]];
b[n_, l_, k_] := b[n, l, k] = Module[{j, pc}, If[n == 0, {1}, pc = {}; For[j = 1, j <= k, j++, pc = zip[Plus, pc, Join[If[jJean-François Alcover, Dec 05 2023, after Alois P. Heinz *)

cp's OEIS Frontend

A386860 The total number of big descents in all parking functions of length n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula