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A322598
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a(n) is the number of unlabeled rank-3 graded lattices with 3 coatoms and n atoms.
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4
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1, 3, 8, 13, 20, 29, 39, 50, 64, 78, 94, 112, 131, 151, 174, 197, 222, 249, 277, 306, 338, 370, 404, 440, 477, 515, 556, 597, 640, 685, 731, 778, 828, 878, 930, 984, 1039, 1095, 1154, 1213, 1274, 1337, 1401, 1466, 1534, 1602, 1672, 1744, 1817
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OFFSET
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1,2
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COMMENTS
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Also number of bicolored graphs, with 3 vertices in the first color class and n in the second, with no isolated vertices, and where any two vertices in one class have at most one common neighbor.
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LINKS
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FORMULA
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a(n) = floor( (3/4)n^2 + (1/3)n + 1/4 ).
G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6.
(End)
a(6*m) = 27*m^2+2*m.
a(6*m+1) = 27*m^2+11*m+1.
a(6*m+2) = 27*m^2+20*m+3.
a(6*m+3) = 27*m^2+29*m+8.
a(6*m+4) = 27*m^2+38*m+13.
a(6*m+5) = 27*m^2+47*m+20.
These imply the conjectured G.f. and recursion.(End)
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EXAMPLE
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a(2)=3: These are the three lattices.
o o o
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o o o o o o o o o
|/ | |/_/| |/ \|
o o o o o o
\ / \ / \ /
o o o
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MAPLE
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seq(floor(3/4*n^2+n/3+1/4), n=1..100); # Robert Israel, Dec 19 2018
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MATHEMATICA
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LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 3, 8, 13, 20, 29}, 50] (* Jean-François Alcover, Dec 29 2018 *)
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PROG
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(PARI) Vec(x*(1 + 2*x + 4*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)) + O(x^50)) \\ Colin Barker, Dec 19 2018
(GAP) List([1..50], n->Int((3/4)*n^2+(1/3)*n+1/4)); # Muniru A Asiru, Dec 20 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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