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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, 0, 1, 0, 5, 0, 7, 14, 0, 19, 52, 0, 34, 173, 48
OFFSET
1,6
COMMENTS
The regions include any holes in the polyominoes.
The first 16 terms were calculated by hand.
EXAMPLE
Triangle begins:
0;
0;
0, 1;
0, 5;
0, 7, 14;
0, 19, 52;
0, 34, 173, 48;
...
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].
.
CROSSREFS
Column 1 gives A000004.
Row lengths give A110654.
Row sums give A380285.
Sequence in context: A084248 A201417 A147666 * A343071 A215892 A200643
KEYWORD
nonn,tabf,more,new
AUTHOR
Omar E. Pol, Jan 18 2025
STATUS
approved