A380283 Irregular triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and width k, and their bounding boxes, n >= 1, 1 <= k <= ceiling(n/2).

0, 0, 0, 1, 0, 5, 0, 7, 14, 0, 19, 52, 0, 34, 173, 48, 0, 74, 503, 384, 0, 134, 1368, 1918, 210, 0, 282, 3642, 7742, 2307, 0, 524, 9552, 26843, 16267, 752, 0, 1064, 24889, 87343, 84789, 11556, 0, 2017, 64200, 272599, 370799, 103336, 2833, 0, 4009, 164826, 838160, 1445347, 678863, 52437

Offset: 1

Author

Omar E. Pol, Jan 18 2025

Keywords

• nonn
• tabf

Comments

The regions include any holes in the polyominoes.

Examples

			Triangle begins:
  0;
  0;
  0,     1;
  0,     5;
  0,     7,      14;
  0,    19,      52;
  0,    34,     173,      48;
  0,    74,     503,     384;
  0,   134,    1368,    1918,      210;
  0,   282,    3642,    7742,     2307;
  0,   524,    9552,   26843,    16267,      752;
  0,  1064,   24889,   87343,    84789,    11556;
  0,  2017,   64200,  272599,   370799,   103336,    2833;
  0,  4009,  164826,  838160,  1445347,   678863,   52437;
  0,  7663,  420373, 2539843,  5240853,  3659815,  560348,  10396;
  0, 15031, 1068181, 7631249, 18171771, 17199831, 4373770, 226716;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there is only one free pentomino of width 1 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,1) = 0.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 2 there are five free pentominoes of width 2 as shown below, and from left to right there are respectively 1, 2, 2, 1, 1 regions between the pentominoes and their bounding boxes, hence the total number of regions is 1 + 2 + 2 + 1 + 1 = 7, so T(5,2) = 7.
   _           _         _
  |_|        _|_|      _|_|      _ _       _ _
  |_|       |_|_|     |_|_|     |_|_|     |_|_|
  |_|_      |_|         |_|     |_|_|     |_|_
  |_|_|     |_|         |_|     |_|       |_|_|
.
For k = 3 there are six free pentominoes of width 3 as shown below, and from left to right there are respectively 3, 2, 1, 2, 4, 2 regions between the pentominoes and their bounding boxes, hence the total number of regions is 3 + 2 + 1 + 2 + 4 + 2 = 14, so T(5,3) = 14.
     _ _     _ _ _     _         _           _       _ _
   _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
    |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
Therefore the 5th row of the triangle is [0, 7, 14].
.
		

Links

• John Mason, Table of n, a(n) for n = 1..90

Crossrefs

Column 1 gives A000004.

Row lengths give A110654.

Row sums give A380285.

Cf. A000105, A379623, A379625, A379626, A379627, A379628, A379637, A380284.

Extensions

More terms from John Mason, Feb 14 2025