A380285 -id:A380285 - cp's OEIS frontend

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).

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

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

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

The first 28 terms were calculated by hand.

			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].
.

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

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.

More terms from John Mason, Feb 14 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

A380286 Number of distinct values of the number of regions between the free polyominoes with n cells and their bounding boxes.

Original entry on oeis.org

1, 1, 2, 3, 5, 5, 5, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 17, 17, 17, 18, 19, 19, 19, 20, 21, 21, 21, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 27, 28, 29, 29, 29, 30, 31, 31, 31, 32, 33, 33, 33, 34, 35, 35, 35, 36, 37, 37, 37

Offset: 1

Omar E. Pol, Jan 24 2025

The regions include any holes in the polyominoes.
From Andrew Howroyd, Mar 01 2025: (Start)
Consider the following sequence of polyominoes for n >= 5:
    O       O         O           O   O       O   O       O   O
  O O O   O O O O   O O O O O   O O O O O   O O O O O   O O O O O O
    O       O         O           O           O   O       O   O
This construction shows how the number of regions between the polyomino and its bounding box can be increased by 2 with the addition of 4 cells. It is also easy to see that any number of fewer holes can also be realized. Moreover, this construction gives the greatest number of regions since except for four corner regions every other region must be bounded on 3 sides by at least one cell separating it from a neighboring region. This leads to a formula for a(n). (End)

			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 there are three distinct values of the number of regions, they are [0, 1, 2], so a(4) = 3.
.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for sequences related to polyominoes.
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Row lengths of A380282.

Cf. A000105, A379627, A379628, A380283, A380284, A380285.

LinearRecurrence[{2, -2, 2, -1}, {1, 1, 2, 3, 5, 5, 5, 6}, 100] (* Paolo Xausa, Mar 02 2025 *)

From Andrew Howroyd, Mar 01 2025: (Start)
a(n) = A004525(n + 4) for n >= 5.
G.f.: x*(1 - x + 2*x^2 - x^3 + 2*x^4 - 2*x^5 + x^6 - x^7)/((1 - x)^2*(1 + x^2)). (End)
E.g.f.: (exp(x)*(4 + x) + sin(x))/2 - 2 - 2*x - x^2 - x^3/6 - x^4/24. - Stefano Spezia, Mar 03 2025

a(8)-a(16) from Pontus von Brömssen, Jan 24 2025
a(17)-a(18) from John Mason, Feb 14 2025
a(19) onwards from Andrew Howroyd, Feb 17 2025

cp's OEIS Frontend

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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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.

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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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A380286 Number of distinct values of the number of regions between the free polyominoes with n cells and their bounding boxes.

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Extensions