OFFSET
1,8
COMMENTS
Here the length is the longer of the two dimensions and the width is the shorter of the two dimensions.
LINKS
John Mason, Table of n, a(n) for n = 1..171 (first 18 rows)
EXAMPLE
Triangle begins:
1;
0, 1;
1, 0, 1;
1, 3, 0, 1;
6, 2, 3, 0, 1;
7, 16, 6, 5, 0, 1;
25, 39, 27, 11, 5, 0, 1;
80, 120, 97, 45, 19, 7, 0, 1;
255, 425, 307, 191, 71, 28, 7, 0, 1;
795, 1565, 1077, 706, 347, 115, 40, 9, 0, 1;
2919, 5217, 4170, 2505, 1454, 574, 171, 53, 9, 0, 1;
10378, 18511, 15164, 10069, 5481, 2740, 919, 257, 69, 11, 0, 1;
...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 0 there are six free pentominoes with length 3 and width 3 as shown below, thus the difference between length and width is 3 - 3 = 0, so T(5,0) = 6.
_ _ _ _ _ _ _ _ _ _
_|_|_| |_|_|_| |_| |_|_ _|_|_ |_|_|
|_|_| |_| |_|_ _ |_|_|_ |_|_|_| |_|_
|_| |_| |_|_|_| |_|_| |_| |_|_|
.
For k = 1 there are two free pentominoes with length 3 and width 2 as shown below, thus the difference between length and width is 3 - 2 = 1, so T(5,1) = 2.
_ _ _ _
|_|_| |_|_|
|_|_| |_|_
|_| |_|_|
.
For k = 2 there are three free pentominoes with length 4 and width 2 as shown below, thus the difference between length and width is 4 - 2 = 2, so T(5,2) = 3.
_ _ _
|_| _|_| _|_|
|_| |_|_| |_|_|
|_|_ |_| |_|
|_|_| |_| |_|
.
For k = 3 there are no free pentominoes whose difference between length and width is 3, so T(5,3) = 0.
For k = 4 there is only one free pentomino with length 5 and width 1 as shown below, thus the difference between length and width is 5 - 1 = 4, so T(5,4) = 1.
_
|_|
|_|
|_|
|_|
|_|
.
Therefore the 5th row of the triangle is [6, 2, 3, 0, 1] and the row sum is A000105(5) = 12.
Note that for n = 6 and k = 1 there are 15 free polyominoes with length 4 and width 3 thus the difference between length and width is 4 - 3 = 1. Also there is a free polyomino with length 3 and width 2 thus the difference between length and width is 3 - 2 = 1, so T(6,1) = 15 + 1 = 16.
.
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Jan 12 2025
EXTENSIONS
Terms a(29) and beyond from Jinyuan Wang, Jan 13 2025
STATUS
approved
