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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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%I #27 Dec 29 2018 07:53:29

%S 1,3,8,13,20,29,39,50,64,78,94,112,131,151,174,197,222,249,277,306,

%T 338,370,404,440,477,515,556,597,640,685,731,778,828,878,930,984,1039,

%U 1095,1154,1213,1274,1337,1401,1466,1534,1602,1672,1744,1817

%N a(n) is the number of unlabeled rank-3 graded lattices with 3 coatoms and n atoms.

%C 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.

%H Jukka Kohonen, <a href="/A322598/b322598.txt">Table of n, a(n) for n = 1..1000</a>

%H J. Kohonen, <a href="http://arxiv.org/abs/1804.03679">Counting graded lattices of rank three that have few coatoms</a>, arXiv:1804.03679 [math.CO] preprint (2018).

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,-1,-1,1).

%F a(n) = floor( (3/4)n^2 + (1/3)n + 1/4 ).

%F From _Colin Barker_, Dec 19 2018: (Start)

%F G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).

%F a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6.

%F (End)

%F From _Robert Israel_, Dec 19 2018: (Start)

%F a(6*m) = 27*m^2+2*m.

%F a(6*m+1) = 27*m^2+11*m+1.

%F a(6*m+2) = 27*m^2+20*m+3.

%F a(6*m+3) = 27*m^2+29*m+8.

%F a(6*m+4) = 27*m^2+38*m+13.

%F a(6*m+5) = 27*m^2+47*m+20.

%F These imply the conjectured G.f. and recursion.(End)

%e a(2)=3: These are the three lattices.

%e o o o

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%p seq(floor(3/4*n^2+n/3+1/4),n=1..100); # _Robert Israel_, Dec 19 2018

%t LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 3, 8, 13, 20, 29}, 50] (* _Jean-François Alcover_, Dec 29 2018 *)

%o (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

%o (GAP) List([1..50],n->Int((3/4)*n^2+(1/3)*n+1/4)); # _Muniru A Asiru_, Dec 20 2018

%Y Third row of A300260.

%Y Next rows are A322599, A322600.

%K nonn,easy

%O 1,2

%A _Jukka Kohonen_, Dec 19 2018