A379623 -id:A379623 - cp's OEIS frontend

A379627 Total area of the bounding boxes of the free polyominoes with n cells.

1, 2, 7, 26, 95, 353, 1387, 5699, 23592, 99020, 415796, 1748722, 7336978, 30741661, 128510698, 536268877, 2233642396, 9288889882

Offset: 1

Omar E. Pol, Jan 07 2025

a(n) is also the total sum of all products of length and width of the free polyominoes with n cells.

			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 bounding boxes is 4 + 6 + 6 + 6 + 4 = 26, so a(4) = 26.
.

Index entries for sequences related to polyominoes.

Cf. A000105, A057766, A379623, A379624, A379628.

a(n) = A057766(n) + A379628(n).

a(7)-a(16) from Pontus von Brömssen, Jan 11 2025

a(17)-a(18) from John Mason, Feb 16 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

Omar E. Pol, Jan 07 2025

The length here is the longer of the two dimensions.

			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.
.

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

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.

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

Omar E. Pol, Jan 12 2025

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

			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.
.

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

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.

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

Omar E. Pol, Jan 07 2025

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

			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.
.

Index entries for sequences related to polyominoes.

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

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

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

Omar E. Pol, Jan 16 2025

The width here is the shorter of the two dimensions.

			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].
.

Index entries for sequences related to polyominoes.

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

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

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

			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].
.

John Mason, Table of n, a(n) for n = 1..90

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

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

Omar E. Pol, Jan 16 2025

The width here is the shorter of the two dimensions.

			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.

Index entries for sequences related to polyominoes.

Row sums of A379637.

Cf. A000105, A379623, A379625, A379627, A379629.

a(15)-a(16) from Pontus von Brömssen, Jan 17 2025

a(17)-a(18) from John Mason, Feb 16 2025

A380282 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells having k regions between the polyominoes and their bounding boxes, n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 4, 5, 1, 1, 2, 6, 18, 7, 2, 1, 13, 50, 34, 10, 2, 25, 144, 146, 50, 2, 2, 48, 402, 574, 240, 18, 1, 2, 97, 1168, 2142, 1120, 122, 4, 1, 201, 3368, 7813, 4920, 738, 32, 3, 420, 9977, 28010, 20946, 4015, 225, 4, 1, 904, 29856, 99610, 86400, 20221, 1561, 37, 1

Offset: 1

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

			Triangle begins:
   1;
   1;
   1,   1;
   2,   1,     2;
   1,   4,     5,     1,     1;
   2,   6,    18,     7,     2;
   1,  13,    50,    34,    10;
   2,  25,   144,   146,    50,     2;
   2,  48,   402,   574,   240,    18,    1;
   2,  97,  1168,  2142,  1120,   122,    4;
   1, 201,  3368,  7813,  4920,   738,   32;
   3, 420,  9977, 28010, 20946,  4015,  225,  4;
   1, 904, 29856, 99610, 86400, 20221, 1561, 37,  1;
   ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 0 there is only one free pentomino having no regions into its bounding box as shown below, so T(5,0) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 1 there are four free pentominoes having only one region into their bounding boxes as shown below, so T(5,1) = 4.
   _
  |_|      _ _     _ _      _
  |_|     |_|_|   |_|_|    |_|
  |_|_    |_|_|   |_|_     |_|_ _
  |_|_|   |_|     |_|_|    |_|_|_|
.
For k = 2 there are five free pentominoes having two regions into their bounding boxes as shown below, so T(5,2) = 5.
     _       _
   _|_|    _|_|    _ _ _    _        _ _
  |_|_|   |_|_|   |_|_|_|  |_|_     |_|_|
  |_|       |_|     |_|    |_|_|_     |_|_
  |_|       |_|     |_|      |_|_|    |_|_|
.
For k = 3 there is only one free pentomino having three regions into its bounding box as shown below, so T(5,3) = 1.
     _ _
   _|_|_|
  |_|_|
    |_|
.
For k = 4 there is only one free pentomino having four regions into its bounding box as shown below, so T(5,4) = 1.
     _
   _|_|_
  |_|_|_|
    |_|
.
Therefore the 5th row of the triangle is [1, 4, 5, 1, 1] and the row sums is A000105(5) = 12.
.

John Mason, Table of n, a(n) for n = 1..116 (first 18 rows (16 rows from _Pontus von Brömssen_))
Index entries for sequences related to polyominoes.

Row sums give A000105.
Row lengths give A380286.
Cf. A379623, A379627, A379628, A379637, A380283, A380284, A380285.
Cf. A038548.

T(n,0) = A038548(n). - Pontus von Brömssen, Jan 24 2025

Terms a(23) and beyond from Pontus von Brömssen, Jan 24 2025

A380285 Total number of regions between the free polyominoes with n cells and their bounding boxes.

Original entry on oeis.org

0, 0, 1, 5, 21, 71, 255, 961, 3630, 13973, 53938, 209641, 815784, 3183642, 12439291, 48686549, 190787588, 748645732

Offset: 1

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

			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:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
From left to right the number of regions between the free tetrominoes and their bounding boxes are respectively 0, 1, 2, 2, 0. Hence the total number of regions is 0 + 1 + 2 + 2 + 0 = 5, so a(4) = 5.
.

Index entries for sequences related to polyominoes.

Row sums of A380283 and of A380284.
Cf. A379628 (total area of the regions).
Cf. A000105, A057766, A379623, A379624, A379625, A379626, A379627, A379629, A379637, A379638, A380282.

a(n) = Sum_{k>0} k*A380282(n,k). - Pontus von Brömssen, Jan 24 2025

a(8)-a(16) from Pontus von Brömssen, Jan 24 2025

a(17)-a(18) from John Mason, Feb 14 2025

A381703 Irregular triangle read by rows in which every row of length A071764(n) lists A(n,w,h) = the number of free polyominoes of size n, width w and height h (for w <= h, and all possible w,h pairs).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 6, 1, 1, 6, 5, 7, 15, 1, 2, 11, 5, 7, 39, 25, 18, 1, 1, 10, 19, 7, 3, 59, 96, 35, 77, 61, 1, 3, 22, 28, 7, 1, 42, 210, 188, 49, 181, 383, 97, 73, 1, 1, 15, 52, 40, 9, 21, 255, 550, 332, 63, 266, 1304, 822, 155, 529, 240, 1, 3, 45, 90, 53, 9, 4, 212, 954, 1231, 529, 81, 251, 2847, 3548, 1551, 220, 2413, 2366, 410, 255

Offset: 1

John Mason, Mar 04 2025

			Triangle begins:
   n
   1:  1
   2:  1
   3:  1  1
   4:  1  1   3
   5:  1  2   3   6
   6:  1  1   6   5   7  15
   7:  1  2  11   5   7  39  25   18
   8:  1  1  10  19   7   3  59   96   35   77   61
   9:  1  3  22  28   7   1  42  210  188   49  181  383    97   73
  10:  1  1  15  52  40   9  21  255  550  332   63  266  1304  822  155  529  240
  ...
Any row contains an irregular array that shows the number of polyominoes having width w and height h. E.g., row 6 contains the array:
  h/w 1  2  3
  1
  2
  3      1  7
  4      6 15
  5      5
  6   1
.
There are 5 polyominoes of size 6 with width 2 and height 5, so A(6,2,5)=5:
.
  OO O  O  O  O
  O  OO O  O  O
  O  O  OO O  OO
  O  O  O  OO  O
  O  O  O   O  O

John Mason, Table of n, a(n) for n = 1..386

Row sums give A000105.
Row lengths give A071764.
Cf. A379623, A379624, A379625, A317186.

More terms from John Mason, Mar 07 2025

cp's OEIS Frontend

A379627 Total area of the bounding boxes of the free polyominoes with n cells.

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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.

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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.

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

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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).

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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).

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

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A380282 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells having k regions between the polyominoes and their bounding boxes, n >= 1, k >= 0.

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A380285 Total number of regions between the free polyominoes with n cells and their bounding boxes.

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