A328433 Number of inversion sequences of length n avoiding the consecutive patterns 011 and 012.
1, 1, 2, 4, 11, 37, 157, 791, 4676, 31490, 238814, 2009074, 18585645, 187366675, 2045016693, 24018394333, 302051731428, 4049206907012, 57642586053512, 868375941780450, 13801973373609889, 230808858283551859, 4051069379668626948, 74459335679007458268
Offset: 0
Keywords
Examples
The a(4)=11 length 4 inversion sequences avoiding the consecutive patterns 011 and 012 are 0000, 0100, 0010, 0020, 0001, 0101, 0021, 0002, 0102, 0003, and 0103.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..464
- 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
# after Alois P. Heinz in A328357 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 = 0 .. n - 1)) end proc: a := n -> b(n, -1, false): seq(a(n), n = 0 .. 24);
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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, 0, n - 1}]]; a[n_] := b[n, -1, False]; a /@ Range[0, 24] (* Jean-François Alcover, Mar 02 2020, after Alois P. Heinz in A328357 *)
Formula
a(n) ~ n! * c * (3^(3/2)/(2*Pi))^n / n^alfa, where alfa = A073016 = Sum_{k>=1} 1/binomial(2*k, k) = 1/3 + 2*Pi/3^(5/2) = 0.73639985871871507790... and c = 2.21611825460684222558745179... - Vaclav Kotesovec, Oct 19 2019
Comments