A135711 -id:A135711 - cp's OEIS frontend

A027709 Minimal perimeter of polyomino with n square cells.

0, 4, 6, 8, 8, 10, 10, 12, 12, 12, 14, 14, 14, 16, 16, 16, 16, 18, 18, 18, 18, 20, 20, 20, 20, 20, 22, 22, 22, 22, 22, 24, 24, 24, 24, 24, 24, 26, 26, 26, 26, 26, 26, 28, 28, 28, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 30, 32, 32, 32, 32, 32, 32, 32, 32, 34, 34, 34, 34, 34, 34

Offset: 0

Jonathan Custance (jevc(AT)atml.co.uk)

			a(5) = 10 because we can arrange 5 squares into 2 rows, with 2 squares in the top row and 3 squares in the bottom row. This shape has perimeter 10, which is minimal for 5 squares.

F. Harary and H. Harborth, Extremal Animals, Journal of Combinatorics, Information & System Sciences, Vol. 1, No 1, 1-8 (1976).
W. C. Yang, Optimal polyform domain decomposition (PhD Dissertation), Computer Sciences Department, University of Wisconsin-Madison, 2003.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Greg Malen, Érika Roldán, and Rosemberg Toalá-Enríquez, Extremal {p, q}-Animals, Ann. Comb. (2023). See Corollary 1.9 at p. 8.
Henri Picciotto, Geometry Labs, Labs 8.1-8.3.
J. Yackel, R. R. Meyer and I. Christou, Minimum-perimeter domain assignment, Mathematical Programming, vol. 78 (1997), pp. 283-303.
Jason R. Zimba, Solution to Perimeter Problem, Jan 23 2015

Cf. A000105, A067628 (analog for triangles), A075777 (analog for cubes).
Cf. A135711.
Number of such polyominoes is in A100092.

a027709 0 = 0
a027709 n = a027434 n * 2  -- Reinhard Zumkeller, Mar 23 2013

[2*Ceiling(2*Sqrt(n)): n in [0..100]]; // Vincenzo Librandi, May 11 2015

interface(quiet=true); for n from 0 to 100 do printf("%d,", 2*ceil(2*sqrt(n))) od;

Table[2*Ceiling[2*Sqrt[n]], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 01 2014 *)

from math import isqrt
def A027709(n): return 1+isqrt((n<<2)-1)<<1 if n else 0 # Chai Wah Wu, Jul 28 2022

a(n) = 2*ceiling(2*sqrt(n)).

a(n) = 2*A027434(n) for n > 0. - Tanya Khovanova, Mar 04 2008

Edited by Winston C. Yang (winston(AT)cs.wisc.edu), Feb 02 2002

A067628 Minimal perimeter of polyiamond with n triangles.

Original entry on oeis.org

0, 3, 4, 5, 6, 7, 6, 7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 11, 12, 11, 12, 13, 12, 13, 12, 13, 14, 13, 14, 15, 14, 15, 14, 15, 16, 15, 16, 15, 16, 17, 16, 17, 16, 17, 18, 17, 18, 17, 18, 19, 18, 19, 18, 19, 18, 19, 20, 19, 20, 19, 20, 21, 20, 21, 20, 21, 20, 21, 22, 21, 22, 21, 22

Offset: 0

Winston C. Yang (winston(AT)cs.wisc.edu), Feb 02 2002

nonn

A polyiamond is a shape made up of n congruent equilateral triangles.

Frank Harary and Heiko Harborth, Extremal animals, J. Combinatorics Information Syst. Sci., 1(1):1-8, 1976.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Greg Malen, Érika Roldán, and Rosemberg Toalá-Enríquez, Extremal {p, q}-Animals, Ann. Comb. (2023). See Corollary 1.9 at p. 8.
Greg Malen and Érika Roldán, Polyiamonds Attaining Extremal Topological Properties, arXiv:1906.08447 [math.CO], 2019.
J. Yackel, R. R. Meyer, and I. Christou, Minimum-perimeter domain assignment, Mathematical Programming, vol. 78 (1997), pp. 283-303.
W. C. Yang and R. R. Meyer, Maximal and minimal polyiamonds, 2002.

Cf. A000105, A000577, A027709 (squares), A057729, A135711.

interface(quiet=true); for n from 0 to 100 do if (1 = 1) then temp1 := ceil(sqrt(6*n)); end if; if ((temp1 mod 2) = (n mod 2)) then temp2 := 0; else temp2 := 1; end if; printf("%d,", temp1 + temp2); od;

a(n)=2*ceil((n+sqrt(6*n))/2)-n; \\ Stefano Spezia, Oct 02 2019

from math import isqrt
def A067628(n): return (c:=isqrt(6*n-1)+1)+((c^n)&1) if n else 0 # Chai Wah Wu, Jul 28 2022

Let c(n) = ceiling(sqrt(6n)). Then a(n) is whichever of c(n) or c(n) + 1 has the same parity as n.

a(n) = 2*ceiling((n + sqrt(6*n))/2) - n (Harary and Harborth, 1976). - Stefano Spezia, Oct 02 2019

A135708 Minimal total number of edges in a polyhex consisting of n hexagonal cells.

Original entry on oeis.org

6, 11, 15, 19, 23, 27, 30, 34, 38, 41, 45, 48, 52, 55, 59, 62, 66, 69, 72, 76, 79, 83, 86, 89, 93, 96, 99, 103, 106, 109, 113, 116, 119, 123, 126, 129, 132, 136, 139, 142, 146, 149, 152, 155, 159, 162, 165, 168, 172, 175, 178, 181, 185, 188, 191, 194, 198, 201, 204, 207, 210

Offset: 1

N. J. A. Sloane, based on an email from Sascha Kurz, Mar 05 2008

nonn

The extremal examples were described by Y. S. Kupitz in 1991.

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.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A135711.

[3*n+Ceiling(Sqrt(12*n-3)): n in [1..65]]; // Vincenzo Librandi, Oct 30 2016

Table[3*n + Ceiling[Sqrt[12*n - 3]], {n,1,25}] (* G. C. Greubel, Oct 29 2016 *)

a(n) = 3*n + ceil(sqrt(12*n-3)); \\ Michel Marcus, Oct 30 2016

from math import isqrt
def A135708(n): return 3*n+1+isqrt(12*n-4) # Chai Wah Wu, Jul 28 2022

a(n) = 3*n + ceiling(sqrt(12*n - 3)). - H. Harborth

2*a(n) - A135711(n) = 6n. - Tanya Khovanova, Mar 07 2008

A075777 Minimal surface area of a rectangular solid with volume n and integer sides.

Original entry on oeis.org

6, 10, 14, 16, 22, 22, 30, 24, 30, 34, 46, 32, 54, 46, 46, 40, 70, 42, 78, 48, 62, 70, 94, 52, 70, 82, 54, 64, 118, 62, 126, 64, 94, 106, 94, 66, 150, 118, 110, 76, 166, 82, 174, 96, 78, 142, 190, 80, 126, 90, 142, 112, 214, 90, 142, 100, 158, 178, 238, 94, 246, 190

Offset: 1

Robert A. Stump (bee_ess107(AT)msn.com), Oct 09 2002

To find minimum surface area (incorrect, see below!), let s1_0 = [n^(1/3)]. Find largest integer s1 such that s1 <= s1_0 and s1 | n. Then let s2_0 = [sqrt(n / s1)]. Find largest integer s2 such that s2 <= s2_0 and s2 | (n / s1). Then s3 = n / (s1 * s2). And minimum surface area a(n) = 2 * (s1 * s2 + s1 * s3 + s2 * s3).

The above algorithm is not correct. Consider n=68: algorithm returns (s1,s2,s3) = (4,1,17) for minimal surface area 178. However, (2,2,17) has a surface area of 144. Consider n=246: algorithm gives (6,1,41) but (3,2,41) has a lower surface area. It appears that the algorithm, by picking s1 to be the highest divisor of n, does not get the chance to choose a pair (s1,s2) that is equal or nearly equal. I wrote a Python script for a new algorithm (posted below). However, it is much slower since it loops through every divisor s1 of n and divisor s2 of a given n/s1 while finding and keeping a minimum surface area. (Ceiling is used to avoid a floating point error on the roots.) - Scott B. Farrar, Sep 29 2015

			a(12) = 32 because side lengths of 2, 2 and 3 will give volume 12 and surface area 32, which is the minimum surface area.

Robert Israel, Table of n, a(n) for n = 1..10000
Dan Meyer, The Math Problem that 1000 Math Teachers Couldn't Solve

Cf. A135711.

N:= 1000: # to get a(1) .. a(N)
A:= Vector(N,1)*6*N;
for p1 from 1 to N do
  for p2 from p1 to floor(N/p1) do
     for p3 from p2 to floor(N/p1/p2) do
        n:= p1*p2*p3;
        a:= 2*(p1*p2 + p2*p3 + p1*p3);
        if a < A[n] then A[n]:= a fi
od od od:
seq(A[i],i=1..N); # Robert Israel, Sep 30 2015

a(n) = {s1_0 = floor(n^(1/3)); s1 = s1_0; while (n % s1 != 0, s1--); s2_0 = floor(sqrt(n/s1)); nds1 = n/s1; s2 = s2_0; while (nds1 % s2 != 0, s2--); s3 = n/(s1*s2); return (2 * (s1 * s2 + s1 * s3 + s2 * s3));} \\ Michel Marcus, Apr 14 2013; script based on 1st algorithm now known to give ok terms up to n=67 only

a(n) = {mins = -1; fordiv(n, x, q = n/x; fordiv(q, y, z = q/y; s = 2*(x*y + y*z +x*z); if (mins ==-1, mins =s, mins = min(mins, s)););); mins;} \\ Michel Marcus, Sep 30 2015

import math
def cubestepdown(n):
    s1_0 = int(math.ceil(n ** (1 / 3.0)))
    minSA = -1
    s1 = s1_0
    while s1>=1:
        while n % s1 > 0:
            s1 = s1 - 1
        s1quot = int(n/s1)
        s2_0 = int(math.ceil(math.sqrt(n/s1)))
        s2 = s2_0
        while s2>=1:
            while s1quot % s2 > 0:
                s2 = s2 - 1
            s3 = int(n / (s1 * s2))
            SA = 2*(s1*s2 + s1*s3 + s2*s3)
            if minSA==-1:
                minSA = SA
            else:
                if SAScott B. Farrar, Sep 29 2015

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A027709 Minimal perimeter of polyomino with n square cells.

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A067628 Minimal perimeter of polyiamond with n triangles.

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A135708 Minimal total number of edges in a polyhex consisting of n hexagonal cells.

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Magma

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A075777 Minimal surface area of a rectangular solid with volume n and integer sides.

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