cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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).

Original entry on oeis.org

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

Views

Author

Jukka Kohonen, Mar 01 2018

Keywords

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

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

Crossrefs

Sum of the d-th antidiagonal is A300221(d+3).
Rows 3-5 are A322598, A322599, A322600.

Programs

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)

A322598 a(n) is the number of unlabeled rank-3 graded lattices with 3 coatoms and n atoms.

Original entry on oeis.org

1, 3, 8, 13, 20, 29, 39, 50, 64, 78, 94, 112, 131, 151, 174, 197, 222, 249, 277, 306, 338, 370, 404, 440, 477, 515, 556, 597, 640, 685, 731, 778, 828, 878, 930, 984, 1039, 1095, 1154, 1213, 1274, 1337, 1401, 1466, 1534, 1602, 1672, 1744, 1817
Offset: 1

Views

Author

Jukka Kohonen, Dec 19 2018

Keywords

Comments

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.

Examples

			a(2)=3: These are the three lattices.
    o          o          o
   /|\        /|\        /|\
  o o o      o o o      o o o
  |/  |      |/_/|      |/ \|
  o   o      o   o      o   o
   \ /        \ /        \ /
    o          o          o

Links

Crossrefs

Third row of A300260.
Next rows are A322599, A322600.

Programs

  • GAP
    List([1..50],n->Int((3/4)*n^2+(1/3)*n+1/4)); # Muniru A Asiru, Dec 20 2018
  • Maple
    seq(floor(3/4*n^2+n/3+1/4),n=1..100); # Robert Israel, Dec 19 2018
  • Mathematica
    LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 3, 8, 13, 20, 29}, 50] (* Jean-François Alcover, Dec 29 2018 *)
  • 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

Formula

a(n) = floor( (3/4)n^2 + (1/3)n + 1/4 ).
From Colin Barker, Dec 19 2018: (Start)
G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6.
(End)
From Robert Israel, Dec 19 2018: (Start)
a(6*m) = 27*m^2+2*m.
a(6*m+1) = 27*m^2+11*m+1.
a(6*m+2) = 27*m^2+20*m+3.
a(6*m+3) = 27*m^2+29*m+8.
a(6*m+4) = 27*m^2+38*m+13.
a(6*m+5) = 27*m^2+47*m+20.
These imply the conjectured G.f. and recursion.(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

Views

Author

Jukka Kohonen, Dec 19 2018

Keywords

Links

Crossrefs

Fifth row of A300260.
Previous rows are A322598, A322599.

Formula

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)
Showing 1-3 of 3 results.

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