

A300260


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).


5



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, 68, 68, 29, 7, 1, 1, 8, 39, 121, 190, 121, 39, 8, 1, 1, 9, 50, 197, 441, 441, 197, 50, 9, 1, 1, 10, 64, 299, 907, 1384, 907, 299, 64, 10, 1
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OFFSET

1,5


COMMENTS

T(n,k) = T(k,n), since taking the duals of the lattices swaps n and k.
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


LINKS



FORMULA

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
T(3,k) = floor( (3/4)k^2 + (1/3)k + 1/4 )
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.
Formulas from (Kohonen 2018).
(End)


EXAMPLE

The table starts:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 3, 4, 5, 6, 7, 8, 9, ...
1, 3, 8, 13, 20, 29, 39, 50, ...
1, 4, 13, 34, 68, 121, 197, ...
1, 5, 20, 68, 190, 441, ...
1, 6, 29, 121, 441, ...
1, 7, 39, 197, ...
1, 8, 50, ...
1, 9, ...
1, ...
...


PROG



CROSSREFS

Sum of the dth antidiagonal is A300221(d+3).


KEYWORD



AUTHOR



STATUS

approved



