A016269 Number of monotone Boolean functions of n variables with 2 mincuts. Also number of Sperner systems with 2 blocks.
1, 9, 55, 285, 1351, 6069, 26335, 111645, 465751, 1921029, 7859215, 31964205, 129442951, 522538389, 2104469695, 8460859965, 33972448951, 136276954149, 546269553775, 2188563950925, 8764714059751, 35090233104309, 140455067207455, 562102681589085, 2249257981411351
Offset: 0
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 292, #8, s(n,2).
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- K. S. Brown, Dedekind's problem
- John Elias, Illustration of Initial Terms: Inverse of the Sierpinski Triangle
- Vladeta Jovovic, Illustration for A016269, A047707, A051112-A051118
- Goran Kilibarda and Vladeta Jovovic, Antichains of Multisets, J. Integer Seqs., Vol. 7, 2004.
- Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
- N. M. Rivière, Recursive formulas on free distributive lattices, J. Combinatorial Theory 5 1968 229--234. MR0231764 (38 #92). - _N. J. A. Sloane_, May 12 2012
- Index entries for sequences related to Boolean functions
- Index entries for linear recurrences with constant coefficients, signature (9,-26,24).
Crossrefs
Programs
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Magma
[(2^n)*(2^n-1)/2-3^n+2^n: n in [2..30]]; // Vincenzo Librandi, Oct 06 2017
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Maple
a:= n-> Stirling2(n+4, 4)-Stirling2(n+3, 4): seq(a(n), n=0..24); # Zerinvary Lajos, Oct 05 2007
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Mathematica
CoefficientList[1/((1-2x)(1-3x)(1-4x)) + O[x]^30, x] (* Jean-François Alcover, Nov 28 2015 *) LinearRecurrence[{9, -26, 24}, {1, 9, 55}, 40] (* Vincenzo Librandi, Oct 06 2017 *)
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PARI
a(n)=(2^n)*(2^n-1)/2-3^n+2^n \\ Charles R Greathouse IV, Mar 22 2016
Formula
G.f.: 1/((1-2*x)*(1-3*x)*(1-4*x)).
a(n-2) = (2^n)*(2^n - 1)/2 - 3^n + 2^n.
From Hieronymus Fischer, Jun 25 2007: (Start)
a(n) = Sum_{0<=i,j,k,<=n, i+j+k=n} 2^i*3^j*4^k.
a(n) = 2^(n+1)*(1+2^(n+2))-3^(n+2). (End)
a(n) = 3*StirlingS2(n+3,4) + StirlingS2(n+3,3). - Ross La Haye, Jan 10 2008
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-2) = f(n,2,2), (n >= 2). - Milan Janjic, Apr 26 2009
E.g.f.: (d^2/dx^2) (exp(2*x)*((exp(x)-1)^2)/2!). See the Sheffer comment given above. - Wolfdieter Lang, Oct 08 2011
a(n) = A006516(n+1) + 3*a(n-1), n>=1, a(0)=1. - Carlos A. Rico A., Jun 22 2019
Comments