A380284 Triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and length k, and their bounding boxes, n >= 1, k >= 1.

0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 16, 5, 0, 0, 0, 14, 48, 9, 0, 0, 0, 12, 145, 89, 9, 0, 0, 0, 3, 354, 453, 138, 13, 0, 0, 0, 0, 608, 1930, 876, 203, 13, 0, 0, 0, 0, 804, 6348, 4930, 1598, 276, 17, 0, 0, 0, 0, 721, 17509, 22575, 10197, 2554, 365, 17, 0, 0, 0, 0, 454, 40067, 91007, 54691, 18984, 3955, 462, 21, 0

Offset: 1

Author

Omar E. Pol, Jan 18 2025

Keywords

• nonn
• tabl

Comments

The regions include any holes in the polyominoes.

The first 28 terms were calculated by hand.

Examples

			Triangle begins:
  0;
  0,  0;
  0,  1,  0;
  0,  0,  5,   0;
  0,  0, 16,   5,  0;
  0,  0, 14,  48,  9,  0;
  0,  0, 12, 145, 89,  9,  0;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there are no free pentominoes of length 1, hence there are no regions, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, hence there are no regions, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 1, 3, 2, 1, 2, 4, 2, hence the total number of regions is 1 + 1 + 3 + 2 + 1 + 2 + 4 + 2 = 16,  so T(5,3) = 16.
   _ _     _ _       _ _     _ _ _     _         _           _       _ _
  |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
  |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
For k = 4 there are three free pentominoes of length 4 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 2, 2, hence the total number of regions is 1 + 2 + 2 = 5,  so T(5,4) = 5.
   _         _       _
  |_|      _|_|    _|_|
  |_|     |_|_|   |_|_|
  |_|_    |_|       |_|
  |_|_|   |_|       |_|
.
For k = 5 there is only one free pentomino of length 5 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,5) = 0.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
Therefore the 5th row of the triangle is [0, 0, 16, 5, 0].
.
		

Links

• John Mason, Table of n, a(n) for n = 1..171 (first 18 rows)
• Index entries for sequences related to polyominoes.

Crossrefs

Column 1 and leading diagonal give A000004.

Column 2 gives A063524.

Row sums give A380285.

Cf. A000105, A379624, A379625, A379627, A379628, A379629, A379638, A380282, A380283.

Extensions

More terms from John Mason, Feb 14 2025