A328357 Number of inversion sequences of length n avoiding the consecutive patterns 000, 001, 011, 012.
1, 1, 2, 1, 4, 6, 36, 117, 804, 4266, 33768, 249144, 2289348, 21353472, 227212824, 2533824900, 30914509212, 398623158096, 5508014798052, 80377645583430, 1242697826967816, 20218588415853480, 346035438765576720, 6206862951272939550, 116518581654518098332
Offset: 0
Keywords
Examples
The a(4)=4 length 4 inversion sequences avoiding the consecutive patterns 000, 001, 011, 012 are 0100, 0101, 0102, 0103. The a(5)=6 length 5 inversion sequences are 01010, 01020, 01021, 01030, 01031, 01032.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..465
- Juan S. Auli and Sergi Elizalde, Consecutive patterns in inversion sequences II: avoiding patterns of relations, arXiv:1906.07365 [math.CO], 2019.
Crossrefs
Programs
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Maple
b:= proc(n, x, t) option remember; `if`(n=0, 1, add( `if`(t and i<=x, 0, b(n-1, i, i<=x)), i=1..n)) end: a:= n-> b(n, 0, false): seq(a(n), n=0..24); # Alois P. Heinz, Oct 14 2019
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Mathematica
b[n_, x_, t_] := b[n, x, t] = If[n == 0, 1, Sum[If[t && i <= x, 0, b[n - 1, i, i <= x]], {i, 1, n}]]; a[n_] := b[n, 0, False]; a /@ Range[0, 24] (* Jean-François Alcover, Feb 25 2020, after Alois P. Heinz *)
Formula
a(n) ~ n! * c * (3^(3/2)/(2*Pi))^n / n^(2*Pi/3^(3/2)), where c = 0.75844492121718325018323312623016463... - Vaclav Kotesovec, Oct 17 2019
Extensions
Terms a(11)..a(16) from Joerg Arndt, Oct 14 2019
a(17)-a(24) from Alois P. Heinz, Oct 14 2019
Comments