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 1-3 of 3 results.

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].
.
		

Crossrefs

Column 1 and leading diagonal give A000004.
Column 2 gives A063524.
Row sums give A380285.

Extensions

More terms from John Mason, Feb 14 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

Views

Author

Omar E. Pol, Jan 18 2025

Keywords

Comments

The regions include 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:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
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.
.
		

Crossrefs

Row sums of A380283 and of A380284.
Cf. A379628 (total area of the regions).

Formula

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

Extensions

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

Views

Author

Omar E. Pol, Jan 24 2025

Keywords

Comments

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)

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:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
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.
.
		

Crossrefs

Programs

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

Formula

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

Extensions

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
Showing 1-3 of 3 results.