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 1-10 of 13 results. Next

A103465 Number of polyominoes that can be formed from n regular unit pentagons (or polypents of order n).

Original entry on oeis.org

1, 1, 2, 7, 25, 118, 551, 2812, 14445, 76092, 403976, 2167116, 11698961, 63544050, 346821209, 1901232614
Offset: 1

Views

Author

Sascha Kurz, Feb 07 2005; definition revised and sequence extended Apr 12 2006 and again Jun 09 2006

Keywords

Comments

Number of 5-polyominoes with n pentagons. A k-polyomino is a non-overlapping union of n regular unit k-gons.
Unlike A051738, these are not anchored polypents but simple polypents. - George Sicherman, Mar 06 2006
Polypents (or 5-polyominoes in Koch and Kurz's terminology) can have holes and this enumeration includes polypents with holes. - George Sicherman, Dec 06 2007

Examples

			a(3)=2 because there are 2 geometrically distinct ways to join 3 regular pentagons edge to edge.
		

Crossrefs

Extensions

Entry revised by N. J. A. Sloane, Jun 18 2006

A103473 Number of polyominoes consisting of 7 regular unit n-gons.

Original entry on oeis.org

24, 108, 551, 333, 558, 1605, 4418, 8350, 17507, 13512, 17775, 30467, 55264, 83252, 134422, 112514, 135175, 195122, 294091, 397852, 566007, 495773, 568602, 751172, 1031920, 1307384, 1729686, 1557663, 1737915, 2169846, 2808616, 3413064
Offset: 3

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Examples

			a(3)=24 because there are 24 polyiamonds consisting of 7 triangles and a(4)=108 because there are 108 polyominoes consisting of 7 squares.
		

Crossrefs

Extensions

More terms from Sascha Kurz, Jun 09 2006

A103470 Number of polyominoes consisting of 4 regular unit n-gons.

Original entry on oeis.org

3, 5, 7, 7, 7, 11, 14, 19, 23, 23, 23, 29, 35, 42, 48, 47, 48, 57, 64, 74, 82, 81, 82, 93, 103, 115, 125, 123, 125, 139, 150, 165, 177, 175, 177, 193, 207, 224, 238, 235, 238, 257, 272, 292, 308, 305, 308, 329, 347, 369, 387, 383, 387, 411, 430, 455, 475, 471, 475
Offset: 3

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Examples

			a(3)=3 because there are 3 polyiamonds consisting of 4 triangles and a(4)=5 because there are 5 polyominoes consisting of 4 squares.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2,-2,2,-2,2,-1,0,0,0,0,0,1,-2,2,-2,2,-2,1},{3,5,7,7,7,11,14,19,23,23,23,29,35,42,48,47,48,57},80] (* Harvey P. Dale, Feb 11 2020 *)
  • PARI
    Vec(-x^3*(x^17 -2*x^16 +3*x^15 -4*x^14 +4*x^13 -4*x^12 +4*x^11 -2*x^10 +2*x^9 -2*x^8 +4*x^7 -x^6 +x^5 +3*x^4 -3*x^3 +3*x^2 -x +3)/((x -1)^3*(x +1)*(x^2 -x +1)^2*(x^2 +1)*(x^2 +x +1)^2*(x^4 -x^2 +1)) + O(x^100)) \\ Colin Barker, Jan 19 2015

Formula

See the link for a formula.
G.f.: -x^3*(x^17 -2*x^16 +3*x^15 -4*x^14 +4*x^13 -4*x^12 +4*x^11 -2*x^10 +2*x^9 -2*x^8 +4*x^7 -x^6 +x^5 +3*x^4 -3*x^3 +3*x^2 -x +3) / ((x -1)^3*(x +1)*(x^2 -x +1)^2*(x^2 +1)*(x^2 +x +1)^2*(x^4 -x^2 +1)). - Colin Barker, Jan 19 2015

A103471 Number of polyominoes consisting of 5 regular unit n-gons.

Original entry on oeis.org

4, 12, 25, 22, 25, 50, 82, 127, 186, 168, 187, 263, 362, 472, 614, 566, 615, 776, 972, 1179, 1437, 1347, 1439, 1711, 2045, 2376, 2786, 2641, 2790, 3204, 3707, 4193, 4790, 4575, 4796, 5380, 6089, 6760, 7578, 7282, 7584, 8373, 9321, 10207, 11282, 10890
Offset: 3

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Examples

			a(3)=4 because there are 4 polyiamonds consisting of 5 triangles and a(4)=12 because there are 12 polyominoes consisting of 5 squares.
		

Crossrefs

A103472 Number of polyominoes consisting of 6 regular unit n-gons.

Original entry on oeis.org

12, 35, 118, 82, 118, 269, 585, 985, 1750, 1438, 1765, 2718, 4336, 6040, 8814, 7678, 8839, 11876, 16410, 20970, 27720, 24998, 27787, 34763, 44687, 54133, 67601, 62252, 67777, 81066, 99420, 116465, 140075, 130711, 140434, 163027, 193587, 221521
Offset: 3

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Examples

			a(3)=12 because there are 12 polyiamonds consisting of 6 triangles and a(4)=35 because there are 35 polyominoes consisting of 6 squares.
		

Crossrefs

A120102 Number of polyominoes consisting of 8 regular unit n-gons.

Original entry on oeis.org

66, 369, 2812, 1448, 2876, 10102, 34838, 73675, 181127, 131801, 185297, 352375, 725869, 1180526, 2104485, 1694978, 2123088, 3291481, 5402087, 7739008, 11832175, 10079003, 11917261, 16624712, 24389611, 32317393, 45260884
Offset: 3

Views

Author

Sascha Kurz, Jun 09 2006

Keywords

Examples

			a(3)=66 because there are 66 polyiamonds consisting of 8 triangles and a(4)=369 because there are 369 polyominoes consisting of 8 squares.
		

Crossrefs

A120103 Number of polyominoes consisting of 9 regular unit n-gons.

Original entry on oeis.org

160, 1285, 14445, 6572, 14982, 65323, 280014, 664411, 1908239, 1314914, 1968684, 4158216, 9707046, 17054708, 33522023, 26019735, 33942901, 56537856, 100952307, 153177526, 251530341, 208524646, 254079408, 374310135, 586169115, 812395658
Offset: 3

Views

Author

Sascha Kurz, Jun 09 2006

Keywords

Examples

			a(3)=160 because there are 160 polyiamonds consisting of 9 triangles and a(4)=1285 because there are 1285 polyominoes consisting of 9 squares.
		

Crossrefs

A103466 Number of polyominoes consisting of n regular unit octagons.

Original entry on oeis.org

1, 1, 3, 11, 50, 269, 1605, 10102, 65323, 430302, 2868320, 19299334, 130807068, 892075515, 6115673262
Offset: 1

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Crossrefs

Extensions

More terms from Sascha Kurz, Jun 09 2006
Definition corrected by John Mason and Sascha Kurz, Sep 20 2020

A103467 Number of polyominoes consisting of n regular unit 9-gons.

Original entry on oeis.org

1, 1, 3, 14, 82, 585, 4418, 34838, 280014, 2285047, 18838395, 156644526, 1311575691
Offset: 1

Views

Author

Sascha Kurz, Feb 07 2005

Keywords

Crossrefs

Extensions

More terms from Sascha Kurz, Jun 09 2006
Definition corrected by John Mason and Sascha Kurz, Sep 20 2020

A260090 Maximum number of kings on an n X n chessboard such that no king attacks more than one other king.

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 21, 26, 33, 40, 48, 56, 65, 74, 85, 96, 108, 120, 133, 146, 161, 176, 192, 208, 225, 242, 261, 280, 300, 320, 341, 362, 385, 408, 432, 456, 481, 506, 533, 560, 588, 616, 645, 674, 705, 736, 768, 800, 833, 866, 901, 936, 972, 1008, 1045
Offset: 1

Views

Author

Rob Pratt, Jul 15 2015

Keywords

Comments

Suggested by a problem involving parking cars in Marx (2015). The Marx problem is slightly different, however, since a solution in her book shows one car that is adjacent to two of its eight neighbors.
Can be formulated as an integer linear programming problem as follows. Define a graph with a node for each cell and an edge for each pair of cells that are a king's move apart. Let binary variable x[i] = 1 if a king appears at node i, and 0 otherwise. The objective is to maximize sum x[i]. Let N[i] be the set of neighbors of node i. To enforce the rule that x[i] = 1 implies sum {j in N[i]} x[j] <= 1, impose the linear constraint sum {j in N[i]} x[j] - 1 <= (|N[i]| - 1) * (1 - x[i]) for each i. - Rob Pratt, Jul 16 2015
An alternative formulation uses constraints x[i] + x[j] + x[k] <= 2 for each forbidden triple of nodes.

Examples

			a(8) = 26:
  XX_XX_XX
  ________
  XX_XX_XX
  ________
  XX_XX_X_
  _______X
  X_X_X___
  X_X_X_XX
a(15) = 85:
  XX_XX_XX_XX_X_X
  ____________X_X
  XX_X_X_X_XX____
  ___X_X_X____X_X
  XX_______XX_X_X
  ___XX_XX_______
  XX_______X_X_XX
  ___X_X_X_X_X___
  XX_X_X_______XX
  _______XX_XX___
  X_X_XX_______XX
  X_X____X_X_X___
  ____XX_X_X_X_XX
  X_X____________
  X_X_XX_XX_XX_XX
		

References

  • Dale Gerdemann et al., Discussions on Sequence Fans Mailing List, July 15 2015.
  • Patricia Marx, Let's Be Less Stupid, Hachette, 2015.

Crossrefs

A103139(n) and A181018(n) are upper bounds.
A260113 is the corresponding sequence for queens.

Programs

  • PARI
    a(n)=if(n==3, 4, (n*(n+2) + if(n%3 == 2, 1, 0) - 3*if(n%6 == 2, 1, 0))/3); \\ Yifan Xie, Mar 22 2025

Formula

Conjecture: For n != 3, a(n) = n(n+2)/3 + [n mod 3 = 2]/3 - [n mod 6 = 2]
Equivalent conjecture for n >= 5: a(n) = a(n-1) + n - A103469(n-2). - Bob Selcoe, Jul 17 2015
See A381749 for proofs of the conjectures. - Yifan Xie, Mar 22 2025

Extensions

More terms from Yifan Xie, Mar 22 2025
Showing 1-10 of 13 results. Next