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A052516
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Number of pairs of sets of cardinality at least 3.
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2
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0, 0, 0, 0, 0, 0, 20, 70, 182, 420, 912, 1914, 3938, 8008, 16172, 32526, 65262, 130764, 261800, 523906, 1048154, 2096688, 4193796, 8388054, 16776614, 33553780, 67108160, 134216970
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OFFSET
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0,7
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COMMENTS
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The number of functions f:[n]->[2] such that f maps at least 3 elements to 1 and at least 3 elements to 2. a(6) = 20 since there are C(6,3) ways to choose the 3 elements of {1,2,3,4,5,6} that f maps to 1. - Dennis P. Walsh, Dec 09 2014
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LINKS
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FORMULA
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E.g.f.: (exp(x) -1)^2 -x*(2+x)*exp(x) +2*x +2*x^2 +x^3 +x^4/4.
(n-1)*a(n+2) + (1-3*n)*a(n+1) + 2*(n+1)*a(n) = 0, a(0) = .. a(5) = 0, a(6) = 20.
G.f.: 2*x^6*(10-15*x+6*x^2)/((1-x)^3*(1-2*x)). - Colin Barker, Feb 19 2012
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MAPLE
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Pairs spec := [S, {S=Prod(B, B), B=Set(Z, 3 <= card)}, labeled]: seq(combstruct[count](spec, size=n), n=0..20);
with (combstruct):ZL:=[S, {S=Sequence(U, card=r), U=Set(Z, card>=3)}, labeled]: seq(count(subs(r=2, ZL), size=m), m=0..20); # Zerinvary Lajos, Mar 09 2007
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MATHEMATICA
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PROG
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(Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (Exp(x) -1-x-x^2/2)^2 )); [0, 0, 0, 0, 0] cat [Factorial(n+5)*b[n]: n in [1..m-6]]; // G. C. Greubel, May 13 2019
(Sage) (2*x^6*(10-15*x+6*x^2)/((1-x)^3*(1-2*x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 13 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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encyclopedia(AT)pommard.inria.fr, Jan 25 2000
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STATUS
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approved
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