%I
%S 1,1,1,1,2,1,1,3,3,1,1,4,8,4,1,1,5,13,13,5,1,1,6,20,34,20,6,1,1,7,29,
%T 68,68,29,7,1,1,8,39,121,190,121,39,8,1,1,9,50,197,441,441,197,50,9,1,
%U 1,10,64,299,907,1384,907,299,64,10,1
%N Table read by antidiagonals: T(n,k) is the number of unlabeled rank3 graded lattices with n coatoms and k atoms (for n,k >= 1).
%C T(n,k) = T(k,n), since taking the duals of the lattices swaps n and k.
%C Number of bicolored graphs, with n and k vertices in the color classes, with no isolated vertices, and where any two vertices in one class have at most one common neighbor.  _Jukka Kohonen_, Mar 08 2018
%H Jukka Kohonen, <a href="/A300260/b300260.txt">Table of n, a(n) for n = 1..210</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).
%F T(2,k) = k. Proof: If the coatoms do not have a common atom, the k atoms can be divided between the two coatoms so that the smaller subset has 1..floor(k/2) atoms. If the coatoms have a common atom, the remaining k1 can be divided so that the smaller subset has 0..floor((k1)/2) atoms. In total this makes k possibilities.  _Jukka Kohonen_, Mar 03 2018
%F From _Jukka Kohonen_, Apr 20 2018 (Start)
%F T(3,k) = floor( (3/4)k^2 + (1/3)k + 1/4 )
%F T(4,k) = (97/144)k^3  (5/6)k^2 + [44/48, 47/48]k + [0, 13, 8, 45, 40, 19, 0, 5, 8, 27, 40, 37]/72. The value of the first bracket depends on whether k is even or odd. The value of the second bracket depends on whether (k mod 12) is 0, 1, 2, ..., 11.
%F Formulas from (Kohonen 2018).
%F (End)
%e The table starts:
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
%e 1, 3, 8, 13, 20, 29, 39, 50, ...
%e 1, 4, 13, 34, 68, 121, 197, ...
%e 1, 5, 20, 68, 190, 441, ...
%e 1, 6, 29, 121, 441, ...
%e 1, 7, 39, 197, ...
%e 1, 8, 50, ...
%e 1, 9, ...
%e 1, ...
%e ...
%o (nauty) genbg Z1 d1 u ${n} ${k} # _Jukka Kohonen_, Mar 08 2018
%Y Sum of the dth antidiagonal is A300221(d+3).
%Y Rows 35 are A322598, A322599, A322600.
%K nonn,tabl
%O 1,5
%A _Jukka Kohonen_, Mar 01 2018
