A379625 Triangle read by rows: T(n,k) is the number of free polyominoes with n cells whose difference between length and width is k, n >= 1, k >= 0.

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

Offset: 1

Author

Omar E. Pol, Jan 12 2025

Keywords

• nonn
• tabl

Comments

Here the length is the longer of the two dimensions and the width is the shorter of the two dimensions.

Examples

			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.
.
		

Links

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

Crossrefs

Row sums give A000105.

Column 1 gives A259088.

Row sums except the column 1 give A259087.

Leading diagonal gives A000012.

Second diagonal gives A000004.

Cf. A109613, A379623, A379624, A379626, A379627, A379628, A379629, A379637, A379638, A380283, A380284.

Extensions

Terms a(29) and beyond from Jinyuan Wang, Jan 13 2025