A322598 -id:A322598 - cp's OEIS frontend

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

Jukka Kohonen, Mar 01 2018

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

			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, ...
  ...

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).

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

Rows 3-5 are A322598, A322599, A322600.

genbg -Z1 -d1 -u ${n} ${k}   # Jukka Kohonen, Mar 08 2018

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)

A322599 a(n) is the number of unlabeled rank-3 graded lattices with 4 coatoms and n atoms.

Original entry on oeis.org

1, 4, 13, 34, 68, 121, 197, 299, 432, 600, 806, 1055, 1352, 1698, 2100, 2561, 3085, 3675, 4338, 5074, 5891, 6790, 7777, 8854, 10029, 11300, 12677, 14160, 15756, 17465, 19297, 21249, 23332, 25544, 27894, 30381, 33016, 35794, 38728, 41815, 45065

Offset: 1

Jukka Kohonen, Dec 19 2018

			a(2)=4: These are the four lattices.
    __o__        __o__       __o__       __o__
   / / \ \      / / \ \     / / \ \     / / \ \
  o o   o o     o o o o     o o o o     o o o o
   \_\ /_/|      \|/ \|      \|/  |     |/   \|
      o   o       o   o       o   o     o     o
       \ /         \ /         \ /       \_ _/
        o           o           o          o

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).

Fourth row of A300260.

Adjacent rows are A322598, A322600.

a(n) = (97/144)n^3 - (5/6)n^2 + [44/48, 47/48]n + [0, 13, 8, -45, 40, -19, 0, -5, 8, -27, 40, -37]/72. The value of the first bracket depends on whether n is even or odd. The value of the second bracket depends on whether (n mod 12) is 0, 1, 2, ..., 11.
Conjectures from Colin Barker, Dec 19 2018: (Start)
G.f.: x*(1 + 3*x + 8*x^2 + 17*x^3 + 21*x^4 + 21*x^5 + 16*x^6 + 7*x^7 + 3*x^8) / ((1 - x)^4*(1 + x)^2*(1 + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - 2*a(n-5) + a(n-8) + a(n-9) - a(n-10) for n>10.
(End)

A322600 a(n) is the number of unlabeled rank-3 graded lattices with 5 coatoms and n atoms.

Original entry on oeis.org

1, 5, 20, 68, 190, 441, 907, 1690, 2916, 4734, 7310, 10836, 15528, 21619, 29365, 39045, 50961, 65434, 82809, 103453, 127751, 156117, 188980, 226794, 270037, 319204, 374813, 437409, 507553, 585831, 672847, 769233, 875637, 992735, 1121218, 1261802

Offset: 1

Jukka Kohonen, Dec 19 2018

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).

Fifth row of A300260.

Previous rows are A322598, A322599.

For n>=3: a(n) = (175/192)n^4 - (3079/480)n^3 + (11771/480)n^2
  - [7268/160, 7273/160]n
  + [33600, 34019, 34072, 33627, 33152, 34915, 33624, 33947, 33472, 33507,
  34520, 34459, 32832, 33827, 34072, 34395, 33344, 34147, 33432, 33947,
  34240, 33699, 33752, 34267, 32832, 34595, 34264, 33627, 33152, 34147,
  34200, 34139, 33472, 33507, 33752, 35035, 33024, 33827, 34072, 33627,
  33920, 34339, 33432, 33947, 33472, 34275, 33944, 34267, 32832, 33827,
  34840, 33819, 33152, 34147, 33432, 34715, 33664, 33507, 33752, 34267] / 960.
  The value of the first bracket depends on whether n is even or odd. The value of the second bracket depends on whether (n mod 60) is 0, 1, 2, ..., 59.
Conjectures from Colin Barker, Dec 20 2018: (Start)
G.f.: x*(1 + 4*x + 14*x^2 + 43*x^3 + 102*x^4 + 184*x^5 + 282*x^6 + 368*x^7 + 411*x^8 + 400*x^9 + 333*x^10 + 237*x^11 + 142*x^12 + 70*x^13 + 26*x^14 + 7*x^15 + x^16) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) + a(n-10) - a(n-13) - a(n-14) + a(n-15) for n>15.
(End)

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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).

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A322599 a(n) is the number of unlabeled rank-3 graded lattices with 4 coatoms and n atoms.

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A322600 a(n) is the number of unlabeled rank-3 graded lattices with 5 coatoms and n atoms.

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