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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Kamil Szpojankowski

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Kamil Szpojankowski has authored 1 sequences.

A331836 Number of noncrossing anti-commutator friendly partitions on {1,2,...,2n}.

Original entry on oeis.org

1, 5, 22, 117, 678, 4162, 26588, 174925, 1177158, 8064854, 56062804, 394443458, 2803490524, 20098913252, 145175116408, 1055463627197, 7717664983366, 56720231324046, 418757618733092, 3104269959560566
Offset: 1

Author

Kamil Szpojankowski, Jan 28 2020

Keywords

Comments

Let n be a positive integer, let sigma be a noncrossing partition on {1,2,...,2n}. Consider the set OuterMax(sigma) := {max(W); W is an outer block of sigma}.
We say that sigma is anti-commutator friendly when it satisfies:
1) OuterMax(sigma) is a subset of the union of {1,3,...,2n-1} and {2n}.
2) For every j in {1, 3,...,2n-1} \ OuterMax(sigma), one has depth(j) is not equal to depth(j + 1), where depth(j) stands for the depth of the block of sigma which contains the number j.
For example for n=1 only partition {{1},{2}} is anti-commutator friendly.
Multiplied by two, gives sequence of Boolean cumulants of ab+ba, for a,b freely independent both distributed (delta_0+delta_2)/2. See M. Fevrier et al. Proposition 6.11.

Crossrefs

Cf. A000257.

Programs

  • Mathematica
    Rest[CoefficientList[Series[1/2 - Sqrt[(1 - 8*x)*(1 - Sqrt[1 - 8*x] - 2*x)/(8*x)], {x, 0, 20}], x]] (* Vaclav Kotesovec, Jan 29 2020 *)
  • PARI
    seq(n)={Vec(1/2 - sqrt((1 - 8*x)*(1-2*x-sqrt(1 - 8*x + O(x^2*x^n)))/(8*x)))} \\ Andrew Howroyd, Jan 28 2020

Formula

G.f.: 1/2 - sqrt( (1 - 8*z)*(1 - 2*z-sqrt(1-8*z))/(8*z) ).
G.f.: 1/2 - sqrt( 1/4 + 3*z - 4*z*g(z) ), where g(z) is the g.f. of A000257.
a(n) ~ sqrt(3) * 2^(3*n-2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 29 2020
D-finite with recurrence n*(2*n+1)*a(n) +2*(-11*n^2+17*n-3)*a(n-1) +24*(-n^2+7*n-8)*a(n-2) +32*(16*n^2-109*n+186)*a(n-3) +256*(n-4)*(2*n-9)*a(n-4)=0. - R. J. Mathar, Mar 06 2022