A322598 a(n) is the number of unlabeled rank-3 graded lattices with 3 coatoms and n atoms.
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
Offset: 1
Examples
a(2)=3: These are the three lattices. o o o /|\ /|\ /|\ o o o o o o o o o |/ | |/_/| |/ \| o o o o o o \ / \ / \ / o o o
Links
- Jukka Kohonen, Table of n, a(n) for n = 1..1000
- J. Kohonen, Counting graded lattices of rank three that have few coatoms, arXiv:1804.03679 [math.CO] preprint (2018).
- Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).
Programs
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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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Maple
seq(floor(3/4*n^2+n/3+1/4),n=1..100); # Robert Israel, Dec 19 2018
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Mathematica
LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 3, 8, 13, 20, 29}, 50] (* Jean-François Alcover, Dec 29 2018 *)
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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
Formula
a(n) = floor( (3/4)n^2 + (1/3)n + 1/4 ).
From Colin Barker, Dec 19 2018: (Start)
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)
From Robert Israel, Dec 19 2018: (Start)
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)
Comments