A379638 Triangle read by rows: T(n,k) is the sum of the lengths of the free polyominoes with n cells and length k, n >= 1, k >= 1.

1, 0, 2, 0, 2, 3, 0, 2, 9, 4, 0, 0, 24, 12, 5, 0, 0, 24, 84, 25, 6, 0, 0, 21, 236, 180, 30, 7, 0, 0, 9, 548, 835, 324, 49, 8, 0, 0, 3, 892, 3345, 1842, 539, 56, 9, 0, 0, 0, 1148, 10445, 9762, 3773, 824, 81, 10, 0, 0, 0, 1020, 27360, 42756, 22659, 6712, 1206, 90, 11, 0, 0, 0, 676, 59595, 165024, 116942, 46808, 11439, 1680, 121, 12

Offset: 1

Author

Omar E. Pol, Jan 16 2025

Keywords

• nonn
• tabl

Comments

The length here is the longer of the two dimensions.

Examples

			Triangle begins:
  1;
  0,  2;
  0,  2,   3;
  0,  2,   9,    4;
  0,  0,  24,   12,     5;
  0,  0,  24,   84,    25,      6;
  0,  0,  21,  236,   180,     30,      7;
  0,  0,   9,  548,   835,    324,     49,     8;
  0,  0,   3,  892,  3345,   1842,    539,    56,     9;
  0,  0,   0, 1148, 10445,   9762,   3773,   824,    81,   10;
  0,  0,   0, 1020, 27360,  42756,  22659,  6712,  1206,   90,   11;
  0,  0,   0,  676, 59595, 165024, 116942, 46808, 11439, 1680,  121,  12;
  ...
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, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, hence the sum of the lengths is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 8*3 = 24, so (5,3) = 24.
   _ _     _ _       _ _     _ _ _     _         _           _       _ _
  |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
  |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
For k = 4 there are three free pentominoes of length 4 as shown below, hence the sum of the lengths is 4 + 4 + 4 = 3*4 = 12, so T(5,4) = 12.
   _         _       _
  |_|      _|_|    _|_|
  |_|     |_|_|   |_|_|
  |_|_    |_|       |_|
  |_|_|   |_|       |_|
.
For k = 5 there is only one free pentomino of length 5 as shown below, so T(5,5) = 5.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
Therefore the 5th row of the triangle is [0, 0, 24, 12, 5].
		

Links

• Index entries for sequences related to polyominoes.

Crossrefs

Row sums give A379629.

Cf. A000105, A379624, A379625, A379627, A379637.

Formula

T(n,k) = k*A379624(n,k).