A379627 -id:A379627 - cp's OEIS frontend

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.

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

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

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions