cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A379623 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 5, 6, 1, 12, 22, 1, 18, 71, 18, 1, 37, 193, 138, 1, 60, 490, 661, 73, 1, 117, 1221, 2547, 769, 1, 200, 3011, 8417, 5189, 255, 1, 379, 7393, 26164, 25920, 3743, 1, 669, 18025, 78074, 108834, 32038, 950, 1, 1250, 43847, 229881, 408217, 201956, 16819
Offset: 1

Views

Author

Omar E. Pol, Jan 07 2025

Keywords

Comments

The width here is the shorter of the two dimensions.

Examples

			Triangle begins:
  1;
  1;
  1,    1;
  1,    4;
  1,    5,     6;
  1,   12,    22;
  1,   18,    71,     18;
  1,   37,   193,    138;
  1,   60,   490,    661,     73;
  1,  117,  1221,   2547,    769;
  1,  200,  3011,   8417,   5189,    255;
  1,  379,  7393,  26164,  25920,   3743;
  1,  669, 18025,  78074, 108834,  32038,   950;
  1, 1250, 43847, 229881, 408217, 201956, 16819;
  ...
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, so T(5,1) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 2 there are five free pentominoes of width 2 as shown below, so T(5,2) = 5.
   _           _         _
  |_|        _|_|      _|_|      _ _       _ _
  |_|       |_|_|     |_|_|     |_|_|     |_|_|
  |_|_      |_|         |_|     |_|_|     |_|_
  |_|_|     |_|         |_|     |_|       |_|_|
.
For k = 3 there are six free pentominoes of width 3 as shown below, so T(5,3) = 6.
     _ _     _ _ _     _         _           _       _ _
   _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
    |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
Therefore the 5th row of the triangle is [1, 5, 6] and the row sum is A000105(5) = 12.
.

Links

Crossrefs

Row sums give A000105(n).
Row lengths give A110654(n).
For free polyominoes of length k see A379624.
Cf. A057051, A352720, A379627, A379628.

Extensions

a(21)-a(56) from Pontus von Brömssen, Jan 11 2025

A379624 Triangle read by rows: T(n,k) is the number of free polyominoes with n cells and length k, n >= 1, k = 1..n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 0, 8, 3, 1, 0, 0, 8, 21, 5, 1, 0, 0, 7, 59, 36, 5, 1, 0, 0, 3, 137, 167, 54, 7, 1, 0, 0, 1, 223, 669, 307, 77, 7, 1, 0, 0, 0, 287, 2089, 1627, 539, 103, 9, 1, 0, 0, 0, 255, 5472, 7126, 3237, 839, 134, 9, 1, 0, 0, 0, 169, 11919, 27504, 16706, 5851, 1271, 168, 11, 1
Offset: 1

Views

Author

Omar E. Pol, Jan 07 2025

Keywords

Comments

The length here is the longer of the two dimensions.

Examples

			Triangle begins:
  1;
  0,  1;
  0,  1,  1;
  0,  1,  3,    1;
  0,  0,  8,    3,     1;
  0,  0,  8,   21,     5,     1;
  0,  0,  7,   59,    36,     5,     1;
  0,  0,  3,  137,   167,    54,     7,    1;
  0,  0,  1,  223,   669,   307,    77,    7,    1;
  0,  0,  0,  287,  2089,  1627,   539,  103,    9,   1;
  0,  0,  0,  255,  5472,  7126,  3237,  839,  134,   9,   1;
  0,  0,  0,  169, 11919, 27504, 16706, 5851, 1271, 168,  11,  1;
  ...
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, so T(5,3) = 8.
   _ _     _ _       _ _     _ _ _     _         _           _       _ _
  |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
  |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
For k = 4 there are three free pentominoes of length 4 as shown below, so T(5,4) = 3.
   _         _       _
  |_|      _|_|    _|_|
  |_|     |_|_|   |_|_|
  |_|_    |_|       |_|
  |_|_|   |_|       |_|
.
For k = 5 there is only one free pentomino of length 5 as shown below, so T(5,5) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
Therefore the 5th row of the triangle is [0, 0, 8, 3, 1] and the row sum is A000105(5) = 12.
.

Links

Crossrefs

Row sums give A000105(n).
Column 1 gives A000007.
Leading diagonal gives A000012.
For free polyominoes of width k see A379623.
Cf. A109613, A379627, A379628.

Extensions

Terms a(37) and beyond from Jinyuan Wang, Jan 08 2025

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.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Jan 12 2025

Keywords

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

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

A379628 Total area between the free polyominoes with n cells and their bounding boxes.

Original entry on oeis.org

0, 0, 1, 6, 35, 143, 631, 2747, 12027, 52470, 227993, 985522, 4235295, 18114067, 77112058, 327000797, 1381807943, 5821692946
Offset: 1

Views

Author

Omar E. Pol, Jan 07 2025

Keywords

Comments

a(n) includes the area of any holes in the polyominoes.

Examples

			Illustration for n = 4:
The free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
The bounding boxes are respectively as shown below:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
  4 x 1   3 X 2   3 X 2   3 X 2   2 X 2
.
The total area of the free tetrominoes is 5*4 = 20.
The total area of the bounding boxes is 4 + 6 + 6 + 6 + 4 = 26.
The total area between the bounding boxes and the free tetrominoes is 26 - 20 = 6, so a(4) = 6.
.

Links

Crossrefs

Cf. A000105, A057766, A379623, A379624, A379627.

Formula

a(n) = A379627(n) - A057766(n).

Extensions

a(7)-a(16) from Pontus von Brömssen, Jan 11 2025
a(17)-a(18) from John Mason, Feb 16 2025

A379637 Irregular triangle read by rows: T(n,k) is the sum of the widths of the free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).

Original entry on oeis.org

1, 1, 1, 2, 1, 8, 1, 10, 18, 1, 24, 66, 1, 36, 213, 72, 1, 74, 579, 552, 1, 120, 1470, 2644, 365, 1, 234, 3663, 10188, 3845, 1, 400, 9033, 33668, 25945, 1530, 1, 758, 22179, 104656, 129600, 22458, 1, 1338, 54075, 312296, 544170, 192228, 6650, 1, 2500, 131541, 919524, 2041085, 1211736, 117733
Offset: 1

Views

Author

Omar E. Pol, Jan 16 2025

Keywords

Comments

The width here is the shorter of the two dimensions.

Examples

			Triangle begins:
  1;
  1;
  1,    2;
  1,    8;
  1,   10,     18;
  1,   24,     66;
  1,   36,    213,     72;
  1,   74,    579,    552;
  1,  120,   1470,   2644,     365;
  1,  234,   3663,  10188,    3845;
  1,  400,   9033,  33668,   25945,    1530;
  1,  758,  22179, 104656,  129600,   22458;
  1, 1338,  54075, 312296,  544170,  192228,   6650;
  1, 2500, 131541, 919524, 2041085, 1211736, 117733;
  ...
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, so T(5,1) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 2 there are five free pentominoes of width 2 as shown below, hence the sum of the widths is 2 + 2 + 2 + 2 + 2 = 5*2 = 10, so T(5,2) = 10.
   _           _         _
  |_|        _|_|      _|_|      _ _       _ _
  |_|       |_|_|     |_|_|     |_|_|     |_|_|
  |_|_      |_|         |_|     |_|_|     |_|_
  |_|_|     |_|         |_|     |_|       |_|_|
.
For k = 3 there are six free pentominoes of width 3 as shown below, hence the sum of the widths is 3 + 3 + 3 + 3 + 3 + 3 = 6*3 = 18, so T(5,3) = 18.
     _ _     _ _ _     _         _           _       _ _
   _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
    |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
Therefore the 5th row of the triangle is [1, 10, 18].
.

Links

Crossrefs

Row lengths give A110654.
Row sums give A379626.
Cf. A000105, A379623, A379625, A379627, A379638.

Formula

T(n,k) = k*A379623(n,k).

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).

Original entry on oeis.org

0, 0, 0, 1, 0, 5, 0, 7, 14, 0, 19, 52, 0, 34, 173, 48, 0, 74, 503, 384, 0, 134, 1368, 1918, 210, 0, 282, 3642, 7742, 2307, 0, 524, 9552, 26843, 16267, 752, 0, 1064, 24889, 87343, 84789, 11556, 0, 2017, 64200, 272599, 370799, 103336, 2833, 0, 4009, 164826, 838160, 1445347, 678863, 52437
Offset: 1

Views

Author

Omar E. Pol, Jan 18 2025

Keywords

Comments

The regions include any holes in the polyominoes.

Examples

			Triangle begins:
  0;
  0;
  0,     1;
  0,     5;
  0,     7,      14;
  0,    19,      52;
  0,    34,     173,      48;
  0,    74,     503,     384;
  0,   134,    1368,    1918,      210;
  0,   282,    3642,    7742,     2307;
  0,   524,    9552,   26843,    16267,      752;
  0,  1064,   24889,   87343,    84789,    11556;
  0,  2017,   64200,  272599,   370799,   103336,    2833;
  0,  4009,  164826,  838160,  1445347,   678863,   52437;
  0,  7663,  420373, 2539843,  5240853,  3659815,  560348,  10396;
  0, 15031, 1068181, 7631249, 18171771, 17199831, 4373770, 226716;
  ...
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].
.

Links

Crossrefs

Column 1 gives A000004.
Row lengths give A110654.
Row sums give A380285.
Cf. A000105, A379623, A379625, A379626, A379627, A379628, A379637, A380284.

Extensions

More terms from John Mason, Feb 14 2025

A380284 Triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and length k, and their bounding boxes, n >= 1, k >= 1.

Original entry on oeis.org

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, 0, 0, 3, 354, 453, 138, 13, 0, 0, 0, 0, 608, 1930, 876, 203, 13, 0, 0, 0, 0, 804, 6348, 4930, 1598, 276, 17, 0, 0, 0, 0, 721, 17509, 22575, 10197, 2554, 365, 17, 0, 0, 0, 0, 454, 40067, 91007, 54691, 18984, 3955, 462, 21, 0
Offset: 1

Views

Author

Omar E. Pol, Jan 18 2025

Keywords

Comments

The regions include any holes in the polyominoes.
The first 28 terms were calculated by hand.

Examples

			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].
.

Links

Crossrefs

Column 1 and leading diagonal give A000004.
Column 2 gives A063524.
Row sums give A380285.
Cf. A000105, A379624, A379625, A379627, A379628, A379629, A379638, A380282, A380283.

Extensions

More terms from John Mason, Feb 14 2025

A379626 Sum of the widths of the free polyominoes with n cells.

Original entry on oeis.org

1, 1, 3, 9, 29, 91, 322, 1206, 4600, 17931, 70577, 279652, 1110758, 4424120, 17647314, 70484576, 281750598, 1127181327
Offset: 1

Views

Author

Omar E. Pol, Jan 16 2025

Keywords

Comments

The width here is the shorter of the two dimensions.

Examples

			For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
There is only one free tetromino of width 1 and there are four free tetrominoes of width 2, hence the sum of the widths is 1 + 2 + 2 + 2 + 2 = 9, so a(4) = 9.

Links

Crossrefs

Row sums of A379637.
Cf. A000105, A379623, A379625, A379627, A379629.

Extensions

a(15)-a(16) from Pontus von Brömssen, Jan 17 2025
a(17)-a(18) from John Mason, Feb 16 2025

A379629 Sum of the lengths of the free polyominoes with n cells.

Original entry on oeis.org

1, 2, 5, 15, 41, 139, 474, 1773, 6686, 26043, 101814, 402297, 1593124, 6332329, 25200575, 100462874, 400908688, 1601541747
Offset: 1

Views

Author

Omar E. Pol, Jan 16 2025

Keywords

Comments

The length here is the longer of the two dimensions.

Examples

			For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
                                     _
             _       _       _      |_|
    _ _     |_|     |_|_    |_|_    |_|
   |_|_|    |_|_    |_|_|   |_|_|   |_|
   |_|_|    |_|_|     |_|   |_|     |_|
.
There are no free tetrominoes of length 1, there is only one free tetromino of length 2, there are three free tetrominoes of length 3 and there is only one free tetromino of length 4, hence the sum of the lengths is 0 + 2 + 3 + 3 + 3 + 4 = 15, so a(4) = 15.

Links

Crossrefs

Row sums of A379638.
Cf. A000105, A379624, A379625, A379626, A379627.

Extensions

a(13)-a(16) from Pontus von Brömssen, Jan 17 2025
a(17)-a(18) (based on A379624 b-file) from Pontus von Brömssen, Feb 21 2025

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.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Jan 16 2025

Keywords

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

Crossrefs

Row sums give A379629.
Cf. A000105, A379624, A379625, A379627, A379637.

Formula

T(n,k) = k*A379624(n,k).
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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