A131482 -id:A131482 - cp's OEIS frontend

A380287 Sum of the perimeters of the free polyominoes with n cells.

4, 6, 16, 48, 142, 472, 1670, 6364, 24604, 97668, 390070, 1570560, 6334644, 25617062, 103669288, 419930444, 1701635046, 6898183050

Offset: 1

Omar E. Pol, Jan 25 2025

The perimeters of any holes are included here.

			Illustration for n = 4:
The free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
From left to right the perimeters are respectively [10, 10, 10, 10, 8] as shown below:
    _
   | |     _       _       _
   | |    | |     | |_    | |_     _ _
   | |    | |_    |_  |   |  _|   |   |
   |_|    |_ _|     |_|   |_|     |_ _|
.
The sum of the perimeters is 10 + 10 + 10 + 10 + 8 = 48, so a(4) = 48.
.

Index entries for sequences related to polyominoes.

Cf. A000105, A027709, A057730, A130622, A131482, A135942, A342243.

See A380575 for another version.

a(n) = Sum_{k=2..n+1} 2*k*A342243(n,k). - Pontus von Brömssen, Jan 27 2025

a(6)-a(18) (using A342243 b-file) from Pontus von Brömssen, Jan 27 2025

A204804 Number of free tree-like convex polyominoes with n cells.

Original entry on oeis.org

1, 1, 2, 4, 10, 21, 49, 104, 227, 468, 976, 1978, 4030, 8095, 16313, 32656, 65503, 130986, 262252, 524330, 1049054, 2097549, 4195633, 8389840

Offset: 1

Joseph O'Rourke, Jan 19 2012

Free: none is a rigid transformation (translation, rotation, reflection or glide reflection) of another. Tree-like: never does a 2x2 subarrangement of squares occur in the shape. So the dual graph is a tree. Convex: every horizontal, or vertical line, meets the shape in either a single segment, or not at all.

			n=1: one square. n=2: a 2x1 rectangle. n=3: a 3x1 rectangle; an L-shape. So the sequence starts: 1,1,2,... Images up to n=8 at the MathOverflow link.

Joseph O'Rourke, MathOverflow Question: Counting restricted polyominoes, July 2011.

Cf. A131482, A359661.

It seems that a(n) = 2^(n-1) + 2^(ceiling(n/2)-1) - b(n), where the g.f. of b(n) is x*(1+x^5+x^6) / ((1-x)^4*(1+x)^2*(1+x^2)), and accordingly this sequence itself is a linear recurrence of order 11 with signature (4,-2,-10,14,-2,-8,14,-13,-2,10,-4); cf. Gerhard Paseman's answer at MathOverflow. - Andrei Zabolotskii, May 21 2025

a(9)-a(16) from Karl Fabian, Jan 22 2012
a(17)-a(18) from John Mason, May 06 2021
a(19)-a(24) from Karl Fabian, May 21 2025

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A380287 Sum of the perimeters of the free polyominoes with n cells.

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