A027709 -id:A027709 - cp's OEIS frontend

A135711 Minimal perimeter of a polyhex with n cells.

6, 10, 12, 14, 16, 18, 18, 20, 22, 22, 24, 24, 26, 26, 28, 28, 30, 30, 30, 32, 32, 34, 34, 34, 36, 36, 36, 38, 38, 38, 40, 40, 40, 42, 42, 42, 42, 44, 44, 44, 46, 46, 46, 46, 48, 48, 48, 48, 50, 50, 50, 50, 52, 52, 52, 52, 54, 54, 54, 54, 54, 56, 56, 56, 56, 58, 58, 58, 58, 58, 60, 60

Offset: 1

Tanya Khovanova, Mar 04 2008

nonn

Y. S. Kupitz, "On the maximal number of appearances of the minimal distance among n points in the plane", in Intuitive geometry: Proceedings of the 3rd international conference held in Szeged, Hungary, 1991; Amsterdam: North-Holland: Colloq. Math. Soc. Janos Bolyai. 63, 217-244.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Li Gan, Stéphane Ouvry, and Alexios P. Polychronakos, Algebraic area enumeration of random walks on the honeycomb lattice, arXiv:2107.10851 [math-ph], 2021.
Greg Malen, Érika Roldán, and Rosemberg Toalá-Enríquez, Extremal {p, q}-Animals, Ann. Comb. (2023). See Corollary 1.9 at p. 8.

Cf. A000228 (number of hexagonal polyominoes (or planar polyhexes) with n cells), A135708.

Analogs for triangles, squares, cubes: A067628, A027709, A075777.

Table[2Ceiling[Sqrt[12n-3]],{n,120}] (* Harvey P. Dale, Dec 29 2019 *)

It is easy to use the formula of Harborth given in A135708 to show that a(n) = 2*ceiling(sqrt(12*n-3)). - Sascha Kurz, Mar 05 2008

2*A135708(n) - a(n) = 6n. - Tanya Khovanova, Mar 07 2008

More terms from N. J. A. Sloane, Mar 05 2008

A262767 Minimum perimeter of a rectangle with area n and integer sides.

Original entry on oeis.org

4, 6, 8, 8, 12, 10, 16, 12, 12, 14, 24, 14, 28, 18, 16, 16, 36, 18, 40, 18, 20, 26, 48, 20, 20, 30, 24, 22, 60, 22, 64, 24, 28, 38, 24, 24, 76, 42, 32, 26, 84, 26, 88, 30, 28, 50, 96, 28, 28, 30, 40, 34, 108, 30, 32, 30, 44, 62, 120, 32

Offset: 1

Tim Cieplowski, Sep 30 2015

a(n) >= A027709(n) = 2*ceiling(2*sqrt(n)). - Dmitry Kamenetsky, Feb 27 2017

			Since 2 * (2 + 3) < 2 * (1+6), a(6) = 10.

Cf. A063655 (semiperimeter).

Two-dimensional equivalent of A075777.

f[n_] := Block[{w = Round@ Sqrt@ n}, While[Mod[n, w] != 0, w--]; 2 (w + Round[n/w])]; Array[f, {60}] (* Michael De Vlieger, Oct 01 2015 *)

a(n) = {local(d); d=divisors(n); 2*(d[(length(d)+1)\2] + d[length(d)\2+1])}
vector(50, n, a(n)) \\ Altug Alkan, Oct 16 2015

def perimeter(area):
    width = round(area ** (1/2))
    while area % width != 0:
        width -= 1
    return 2*(width + round(area/width))

a(n) = 2*A063655(n). - Michel Marcus, Oct 01 2015

A331207 Minimum number of matchsticks sufficient to form the edges of exactly n squares of any size >= 1.

Original entry on oeis.org

0, 4, 7, 10, 13, 12, 15, 18, 17, 20, 23, 22, 25, 28, 24, 27, 30, 29, 32, 35, 31, 34, 37, 36, 39, 42, 38, 41, 44, 43, 40, 43, 46, 45

Offset: 0

John King, Jan 12 2020

John King, diagram for A331207 & A331208

Cf. A027709, A078633, A331208.

A331208 The minimum perimeter for exactly n matchstick squares of size >= 1.

Original entry on oeis.org

0, 4, 6, 8, 10, 8, 10, 12, 10, 12, 14, 12, 14, 16, 12, 14, 16, 14, 16, 18, 14, 16, 18, 16, 18, 20, 16, 18, 18, 18, 16, 18, 20

Offset: 0

John King, Jan 12 2020

John King, Graphic showing initial terms and growth for A331207 & A331208

Cf. A027709, A331207, A078633.

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

Original entry on oeis.org

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

A384428 a(n) is the minimal area of a polyomino without holes having a product of edge lengths equal to n, or 0 if no solution is possible.

Original entry on oeis.org

1, 0, 4, 2, 7, 0, 10, 5, 3, 0, 16, 4, 19, 0, 6, 4, 25, 0, 28, 6, 9, 0, 34, 5, 5, 0, 6, 9, 43, 0, 46, 6, 15, 0, 8, 5, 55, 0, 18, 6, 61, 0, 64, 15, 8, 0, 70, 6, 7, 0

Offset: 1

Gordon Hamilton, May 28 2025

Good sequence for elementary school students learning multiplication.
If p is the largest prime factor dividing n, then a(n) >= p because there needs to be at least one edge of length k*p for some k>=1.
a(51) > 21. - Sean A. Irvine, Jun 13 2025

			a(36)=5 because the V pentomino is the smallest polyomino whose edges multiply together to give 36. The edges of the V pentomino are: 3,3,2,2,1,1.
   XXX
   X
   X
a(45)=8 because of the following polyomino with edges 5,3,3,1,1,1,1,1.
   XXXXX
   XX
   X

Sean A. Irvine, Java program (github)

Cf. A000104, A027709.

a(4*n+2) = 0.
a(p) = p + (p-1)/2 for any odd prime p.
a(p^2) = p for any prime p.

a(33)-a(50) from Sean A. Irvine, Jun 13 2025

A283056 Size of the smallest polyomino that admits a hole of size n.

Original entry on oeis.org

0, 7, 9, 11, 11, 13, 13

Offset: 0

Dmitry Kamenetsky, Feb 27 2017

The task here is to surround a hole of size n with the least number of squares. The hole is another polyomino, so we can obtain a lower bound using A027709: minimal perimeter of polyomino with n square cells. We need an extra 3 (or more) diagonal cells to surround any hole. Hence a(n) >= A027709(n) + 3 = 2*ceiling(2*sqrt(n)) + 3.
For rectangular holes we can obtain an upper bound using A262767: minimum perimeter of a rectangle with area n and integer sides. Hence a(n) <= A262767(n) + 3.
Perhaps a(n) is actually equal to A027709(n)+3?

			For n=1, we have a single square hole, so a(1)=7.
For n=2, we have a domino hole, so a(2)=9.
For n=3, we can use either an L or V tromino hole, so a(3)=11.
For n=4, we use the square tetromino hole, so a(4)=11.
For n=5, we use the P pentomino hole, so a(5)=13.
For n=6, we use the 2 X 3 rectangle hole, so a(6)=13.

Cf. A027709.

cp's OEIS Frontend

A135711 Minimal perimeter of a polyhex with n cells.

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A262767 Minimum perimeter of a rectangle with area n and integer sides.

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A331207 Minimum number of matchsticks sufficient to form the edges of exactly n squares of any size >= 1.

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A331208 The minimum perimeter for exactly n matchstick squares of size >= 1.

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

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A384428 a(n) is the minimal area of a polyomino without holes having a product of edge lengths equal to n, or 0 if no solution is possible.

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A283056 Size of the smallest polyomino that admits a hole of size n.

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