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.

A027709 Minimal perimeter of polyomino with n square cells.

Original entry on oeis.org

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

Views

Author

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

Keywords

Examples

			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.

References

  • 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.

Links

Crossrefs

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

Programs

  • Haskell
    a027709 0 = 0
a027709 n = a027434 n * 2  -- Reinhard Zumkeller, Mar 23 2013
  • Magma
    [2*Ceiling(2*Sqrt(n)): n in [0..100]]; // Vincenzo Librandi, May 11 2015
  • Maple
    interface(quiet=true); for n from 0 to 100 do printf("%d,", 2*ceil(2*sqrt(n))) od;
  • Mathematica
    Table[2*Ceiling[2*Sqrt[n]], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 01 2014 *)
  • Python
    from math import isqrt
def A027709(n): return 1+isqrt((n<<2)-1)<<1 if n else 0 # Chai Wah Wu, Jul 28 2022

Formula

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

Extensions

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

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

Views

Author

Tanya Khovanova, Mar 04 2008

Keywords

References

  • 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.

Links

Crossrefs

Cf. A000228 (number of hexagonal polyominoes (or planar polyhexes) with n cells), A135708.
Analogs for triangles, squares, cubes: A067628, A027709, A075777.

Programs

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

Formula

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

Extensions

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

Views

Author

Tim Cieplowski, Sep 30 2015

Keywords

Comments

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

Examples

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

Crossrefs

Cf. A063655 (semiperimeter).
Two-dimensional equivalent of A075777.

Programs

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

Formula

a(n) = 2*A063655(n). - Michel Marcus, Oct 01 2015
Showing 1-3 of 3 results.

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