A300260 Table read by antidiagonals: T(n,k) is the number of unlabeled rank-3 graded lattices with n coatoms and k atoms (for n,k >= 1).
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
Offset: 1
Examples
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, ... ...
Links
- Jukka Kohonen, Table of n, a(n) for n = 1..210
- J. Kohonen, Counting graded lattices of rank three that have few coatoms, arXiv:1804.03679 [math.CO] preprint (2018).
Programs
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nauty
genbg -Z1 -d1 -u ${n} ${k} # Jukka Kohonen, Mar 08 2018
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 k-1 can be divided so that the smaller subset has 0..floor((k-1)/2) atoms. In total this makes k possibilities. - Jukka Kohonen, Mar 03 2018
From Jukka Kohonen, Apr 20 2018 (Start)
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)
Comments