A380282 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells having k regions between the polyominoes and their bounding boxes, n >= 1, k >= 0.

1, 1, 1, 1, 2, 1, 2, 1, 4, 5, 1, 1, 2, 6, 18, 7, 2, 1, 13, 50, 34, 10, 2, 25, 144, 146, 50, 2, 2, 48, 402, 574, 240, 18, 1, 2, 97, 1168, 2142, 1120, 122, 4, 1, 201, 3368, 7813, 4920, 738, 32, 3, 420, 9977, 28010, 20946, 4015, 225, 4, 1, 904, 29856, 99610, 86400, 20221, 1561, 37, 1

Offset: 1

Author

Omar E. Pol, Jan 18 2025

Keywords

• nonn
• tabf

Comments

The regions include any holes in the polyominoes.

Examples

			Triangle begins:
   1;
   1;
   1,   1;
   2,   1,     2;
   1,   4,     5,     1,     1;
   2,   6,    18,     7,     2;
   1,  13,    50,    34,    10;
   2,  25,   144,   146,    50,     2;
   2,  48,   402,   574,   240,    18,    1;
   2,  97,  1168,  2142,  1120,   122,    4;
   1, 201,  3368,  7813,  4920,   738,   32;
   3, 420,  9977, 28010, 20946,  4015,  225,  4;
   1, 904, 29856, 99610, 86400, 20221, 1561, 37,  1;
   ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 0 there is only one free pentomino having no regions into its bounding box as shown below, so T(5,0) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 1 there are four free pentominoes having only one region into their bounding boxes as shown below, so T(5,1) = 4.
   _
  |_|      _ _     _ _      _
  |_|     |_|_|   |_|_|    |_|
  |_|_    |_|_|   |_|_     |_|_ _
  |_|_|   |_|     |_|_|    |_|_|_|
.
For k = 2 there are five free pentominoes having two regions into their bounding boxes as shown below, so T(5,2) = 5.
     _       _
   _|_|    _|_|    _ _ _    _        _ _
  |_|_|   |_|_|   |_|_|_|  |_|_     |_|_|
  |_|       |_|     |_|    |_|_|_     |_|_
  |_|       |_|     |_|      |_|_|    |_|_|
.
For k = 3 there is only one free pentomino having three regions into its bounding box as shown below, so T(5,3) = 1.
     _ _
   _|_|_|
  |_|_|
    |_|
.
For k = 4 there is only one free pentomino having four regions into its bounding box as shown below, so T(5,4) = 1.
     _
   _|_|_
  |_|_|_|
    |_|
.
Therefore the 5th row of the triangle is [1, 4, 5, 1, 1] and the row sums is A000105(5) = 12.
.
		

Links

• John Mason, Table of n, a(n) for n = 1..116 (first 18 rows (16 rows from _Pontus von Brömssen_))
• Index entries for sequences related to polyominoes.

Crossrefs

Row sums give A000105.

Row lengths give A380286.

Cf. A379623, A379627, A379628, A379637, A380283, A380284, A380285.

Cf. A038548.

Formula

T(n,0) = A038548(n). - Pontus von Brömssen, Jan 24 2025

Extensions

Terms a(23) and beyond from Pontus von Brömssen, Jan 24 2025