A067628 -id:A067628 - 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

A069813 Maximum number of triangles in polyiamond with perimeter n.

Original entry on oeis.org

1, 2, 3, 6, 7, 10, 13, 16, 19, 24, 27, 32, 37, 42, 47, 54, 59, 66, 73, 80, 87, 96, 103, 112, 121, 130, 139, 150, 159, 170, 181, 192, 203, 216, 227, 240, 253, 266, 279, 294, 307, 322, 337, 352, 367, 384, 399, 416, 433, 450, 467, 486, 503, 522, 541, 560, 579

Offset: 3

Winston C. Yang (winston(AT)cs.wisc.edu), Apr 30 2002

			a(10) = 16 because the maximum number of triangles in a polyiamond of perimeter 10 is 16.

Colin Barker, Table of n, a(n) for n = 3..1000
W. C. Yang, R. R. Meyer, Maximal and minimal polyiamonds, 2002.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Cf. A000105, A000577, A027709, A030511 (bisection), A057729, A067628.

R:=PowerSeriesRing(Integers(), 65); Coefficients(R!( x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)))); // Marius A. Burtea, Jan 03 2020

A069813 := proc(n)
    round(n^2/6) ;
    if modp(n,6) <> 0 then
        %-1 ;
    else
        % ;
    end if;
end proc: # R. J. Mathar, Jul 14 2015

LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 2, 3, 6, 7, 10}, 60] (* Jean-François Alcover, Jan 03 2020 *)

a(n) = round(n^2/6) - (n % 6 != 0) \\ Michel Marcus, Jul 17 2013

Vec(x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)) + O(x^60)) \\ Colin Barker, Jan 19 2015

a(n) = round(n^2/6) - (0 if n = 0 mod 6, 1 else) = A056829(n)-A097325(n).
From Colin Barker, Jan 18 2015: (Start)
a(n) = round((-25 + 9*(-1)^n + 8*exp(-2/3*i*n*Pi) + 8*exp((2*i*n*Pi)/3) + 6*n^2)/36), where i=sqrt(-1).
G.f.: x^3*(1+x-x^2)*(1+x^2) / ((1-x)^3*(1+x)*(1+x+x^2)). (End)
a(n) = A001399(n-3) + A001399(n-4) + A001399(n-6) - A001399(n-7). - R. J. Mathar, Jul 14 2015

A135711 Minimal perimeter of a polyhex with n cells.

Original entry on oeis.org

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

A008749 Expansion of (1+x^6)/((1-x)(1-x^2)(1-x^3)).

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 9, 12, 15, 18, 21, 26, 29, 34, 39, 44, 49, 56, 61, 68, 75, 82, 89, 98, 105, 114, 123, 132, 141, 152, 161, 172, 183, 194, 205, 218, 229, 242, 255, 268, 281, 296, 309, 324, 339, 354, 369

Offset: 0

N. J. A. Sloane

nonn

Conjecture: For n >= 1, A067628(a(n+2)) appears for the first time in A067628. Equivalently, A067628(a(n+2)) is the first T such that the minimal perimeter of polyiamonds of T triangles is a(n+2). - Winston C. Yang (winston(AT)cs.wisc.edu), Feb 05 2002

			Let n = 8. Then a(n+2) = a(10) = 18. Note A067628(18) = 12 and is the first appearance of 12 in A067628. Equivalently, 12 is the first T such that the min perimeter of polyiamonds of T triangles is 18.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Cf. A067628.

a:=[1,1,2,3,4,5];; for n in [7..60] do a[n]:=a[n-1]+a[n-2]-a[n-4] -a[n-5]+a[n-6]; od; a; # G. C. Greubel, Aug 03 2019

R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (1+x^6)/((1-x)*(1-x^2)*(1-x^3)) )); // G. C. Greubel, Aug 03 2019

CoefficientList[Series[(1+x^6)/((1-x)*(1-x^2)*(1-x^3)), {x,0,60}], x] (* G. C. Greubel, Aug 03 2019 *)

my(x='x+O('x^60)); Vec((1+x^6)/((1-x)*(1-x^2)*(1-x^3))) \\ G. C. Greubel, Aug 03 2019

((1+x^6)/((1-x)*(1-x^2)*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

Conjecture: Let b(n>=0) = (0, 1, 1, 1, 1, 3, 1, 3, 3, 3, 3, 5, 3, 5, 5, 5, 5, 7, 3, ...). Equivalently, let b(0) = 0, b(n>=1) = 2*floor((n-1)/6) + 1 + (2 if n+1=0 mod 6; 0 else). Then a(0) = 1, a(n>=1) = a(n-1) + b(n-1). - Winston C. Yang (winston(AT)cs.wisc.edu), Feb 05 2002

a(n) = (47 + 6*n^2 + 9*(-1)^n + 8*A099837(n+3))/36, n>0. - R. J. Mathar, Jun 24 2009

A137228 Minimal total number of edges in a polyiamond consisting of n triangular cells.

Original entry on oeis.org

3, 5, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 24, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 42, 44, 46, 47, 49, 51, 52, 54, 55, 57, 59, 60, 62, 63, 65, 67, 68, 70, 71, 73, 75, 76, 78, 79, 81, 83, 84, 86, 87, 89, 90, 92, 94, 95, 97, 98, 100, 102, 103, 105, 106, 108, 109, 111, 113

Offset: 1

Tanya Khovanova, Mar 07 2008

nonn

Cf. A135708, A067628, A000577.

a(n) = (3n + A067628(n))/2.

A290648 a(n) is the smallest number of faces of the triangular lattice required to enclose an area consisting of exactly n faces.

Original entry on oeis.org

6, 12, 14, 16, 18, 19, 18, 20, 22, 23, 22

Offset: 0

Christian Majenz, Aug 08 2017

Triangular version of A235382.

Conjecture: a(n) = 2*A067628(n) + 6.

A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

Original entry on oeis.org

9, 14, 19, 23, 27, 31, 35, 39, 43, 47, 51, 54, 58, 62, 65, 69, 73, 76, 80, 83, 87, 90, 94, 97, 101, 104, 108, 111, 115, 118, 122, 125, 129, 132, 135, 139, 142, 146, 149, 152, 156, 159, 163, 166, 169, 173, 176, 179, 183, 186, 189, 193, 196, 199, 203, 206, 209, 213

Offset: 1

Stefano Spezia, Sep 29 2019

nonn

For n > 0, it is easy to prove that k(n) = floor((3 + sqrt(3*(3+8*n)))/6) is the unique integer that satisfies the inequalities 3*binomial(k,2) <= n <= 3*binomial(k+1,2) of Theorem 1.1 in Malen and Roldán.

Proof: solving in k the above inequalities for n > 0, one gets that x - 1 <= k <= x, where x = (3 + sqrt(3*(3+8*n)))/6. Since 3*(3+8*n) is never a perfect square, it follows that x is not an integer and k = floor(x). QED.

Greg Malen and Érika Roldán, Polyiamonds Attaining Extremal Topological Properties, arXiv:1906.08447 [math.CO], 2019.

Cf. A000217, A000577, A067628, A069813, A070764, A161680.

k:=n->floor((3+sqrt(3*(3+8*n)))/6): a:=n->3*(n+k(n))+1+ceil(2*n/k(n)): seq(a(n), n = 1 .. 58)

k[n_]:=Floor[(3+Sqrt[3*(3+8n)])/6]; a[n_]:=3(n+k[n])+1+Ceiling[2n/k[n]]; Array[a,58]

a(n) = 3*(n + k(n)) + 1 + ceiling(2*n/k(n)), where k(n) = floor((3 + sqrt(3*(3+8*n)))/6).

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A027709 Minimal perimeter of polyomino with n square cells.

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A069813 Maximum number of triangles in polyiamond with perimeter n.

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A135711 Minimal perimeter of a polyhex with n cells.

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A008749 Expansion of (1+x^6)/((1-x)*(1-x^2)*(1-x^3)).

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A137228 Minimal total number of edges in a polyiamond consisting of n triangular cells.

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A290648 a(n) is the smallest number of faces of the triangular lattice required to enclose an area consisting of exactly n faces.

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A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

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A008749 Expansion of (1+x^6)/((1-x)(1-x^2)(1-x^3)).