OFFSET
1,9
COMMENTS
The regions include any holes in the polyominoes.
The first 28 terms were calculated by hand.
EXAMPLE
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].
.
CROSSREFS
KEYWORD
AUTHOR
Omar E. Pol, Jan 18 2025
STATUS
approved