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 1-5 of 5 results.

A084618 Maximum number of circles of area 1 that can be packed in a circle of area n.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 4, 5, 7, 7, 8, 8, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 19, 19, 19, 20, 21, 21, 22, 23, 24, 24, 26, 27, 27, 28, 29, 30, 31, 31, 32, 33, 34, 35, 37, 37, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 45, 46, 47, 48, 48, 49, 50, 51, 52, 53, 55, 55, 55, 56, 57, 58, 59
Offset: 1

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Author

Hugo Pfoertner, Jun 01 2003

Keywords

Comments

For most values of n these are only conjectures, supported by numerical results.

Examples

			a(4)=2 because a circle of area 4 is the smallest one covering two circles of area 1.
a(9)=7 is the arrangement of 6 circles closely packed around another circle. This arrangement fits into a circle that has 3*radius of smaller circles and thus 9*their area.
		

References

  • For list of references given by E. Specht, see corresponding link.

Crossrefs

Equivalent sequences for packing into a square: A337020, and equilateral triangle: A337019.

A269110 Numbers of unit circles packed in a triangle of smallest area admitting an equilateral triangle solution.

Original entry on oeis.org

1, 3, 5, 6, 9, 10, 14, 15, 20, 21
Offset: 1

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Author

Leonid Broukhis, Feb 19 2016

Keywords

Crossrefs

May be related to A117142.
For packing into equilateral triangles, see A337019.

A337020 Maximum number of circles with unit area that can be packed into a square with an area of n.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 4, 5, 5, 6, 8, 9, 9, 9, 10, 11, 12, 13, 13, 15, 16, 16, 16, 18, 18, 20, 20, 21, 22, 22, 24, 25, 25, 25, 27, 27, 28, 30, 30
Offset: 1

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Author

Ya-Ping Lu, Nov 06 2020

Keywords

Comments

Terms beyond a(n) = 30 (n = 38 & 39) except a(n) = 36 are conjectures supported by numerical results (see Packomania in the links) and terms for n from 40 through 70 are: 31, 31, 33, 33, 34, 35, 36, 36, 38, 39, 39, 40, 41, 42, 42, 43, 44, 44, 46, 46, 47, 48, 49, 50, 52, 52, 53, 53, 54, 55, 56.
The packing density, a(n)/n, approaches sqrt(3)*Pi/6 as n tends to infinity.
References for the known optimal packings are given in Table 1.2 on page 10 and the bibliography on pages 219-225 of the book by Szabo et al. (see References).

References

  • P. G. Szabo, M. Cs. Markot, T. Csendes, E. Specht, L. G. Casado, and I. Garcia, New Approaches to Circle Packing in a Square, Springer, 2007.

Crossrefs

A343262 a(n) is the number of edges of a regular polygon P with the property that packing n nonoverlapping equal circles inside P, arranged in a configuration with dihedral symmetry D_{2m} with m >= 3, maximizes the packing density.

Original entry on oeis.org

3, 4, 5, 3, 6, 7, 4, 3, 5, 6, 6, 7, 3, 4, 4, 6, 6, 4, 3
Offset: 3

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Author

Ya-Ping Lu, Apr 09 2021

Keywords

Comments

Numbers of dihedral symmetries D_{2m} (m >= 3) that n nonoverlapping equal circles possess are given in A343005. The regular polygon is a circle for n=1 and a square for n=2. However, as the symmetry types, O(2) for one circle and D_{4} for two circles, are not D_{2m} with m >= 3, the index of the sequence starts at n = 3.
It can be shown that a(n) <= n and a(n) = k*m/2, where m is the order of a dihedral symmetry of n-circle packing configurations and k is a positive integer.

Examples

			For n=3, 3-circle configurations possess one dihedral symmetry D_{6}, or m = 3. Since a(n) must be <= 3 and also a multiple of m, a(n) = 3.
For n = 16, 16-circle configurations have 6 D_{2m} symmetries with m >= 3.
Packing densities are for
m = 16: Pi/(2+2*csc(Pi/8)) = 0.43474+,
m = 15: (8*Pi/15)/(1+csc(2*Pi/15)) = 0.48445+,
m =  8: 4*sqrt(2)*Pi/(1+sqrt(2)+sqrt(3)+sqrt(4-2*sqrt(2)))^2 = 0.65004+,
m =  5: (16*Pi/5)*(7-3*sqrt(5))/sqrt(10+2*sqrt(5)) = 0.77110+,
m =  4: Pi/4 = 0.78539+,
m =  3: 8*Pi/(12+13*sqrt(3)) = 0.72813+.
The highest packing density is achieved at m = 4, or a(16) = 4.
Symmetry type (S) of n-circle configuration giving the highest packing density and the corresponding number of edges (N) of the regular polygon and packing density are given below. The packing configurations are illustrated in the Links.
   n       S      N      Packing density
------  --------  --  -------------------------------------------------------------
   3      D_{6}   3   Pi/(2+4/sqrt(3))                                   = 0.72900+
  4,9,16  D_{8}   4   Pi/4                                               = 0.78539+
   5      D_{10}  5   Pi/(2+8/sqrt(10+2*sqrt(5)))                        = 0.76569+
   6      D_{6}   3   6*Pi/(12+7*sqrt(3))                                = 0.78134+
   7      D_{12}  6   7*Pi/(12+8*sqrt(3))                                = 0.85051+
   8      D_{14}  7   4*Pi/(7+7/sin(2*Pi/7))                             = 0.78769+
  10      D_{6}   3   5*Pi/(9+6*sqrt(3))                                 = 0.81001+
  11      D_{10}  5   (22*Pi/25)/sqrt(10+2*sqrt(5))                      = 0.72671+
  12      D_{6}   6   6*Pi/(12+7*sqrt(3))                                = 0.78134+
  13      D_{12}  6   13*sqrt(3)*Pi/96                                   = 0.73685+
  14      D_{14}  7   4*Pi/(sin(2*Pi/7)*(sqrt(3)+cot(Pi/7)+sec(Pi/7))^2) = 0.66440+
  15      D_{6}   3   15*Pi/(24+19*sqrt(3))                              = 0.82805+
  17      D_{8}   4   (17*Pi/4)/(7+3*sqrt(2)+3*sqrt(3)+sqrt(6))          = 0.70688+
  18      D_{12}  6   9*Pi/(12+13*sqrt(3))                               = 0.81915+
  19      D_{12}  6   19*Pi/(24+26*sqrt(3))                              = 0.86465+
  20      D_{8}   4   20*Pi/(2+sqrt(2)+2*sqrt(3)+sqrt(6))^2              = 0.72213+
  21      D_{6}   3   21*Pi/(30+28*sqrt(3))                              = 0.84045+
		

Crossrefs

A342843 a(n) is the number of edges of the regular polygon such that packing n nonoverlapping equal circles inside the regular polygon gives the highest packing density. a(n) = 0 if such a regular polygon is a circle.

Original entry on oeis.org

0, 4, 3, 4, 5, 3, 6, 7, 4, 3, 9, 6, 10, 6, 3, 4
Offset: 1

Views

Author

Ya-Ping Lu, Apr 12 2021

Keywords

Comments

Terms for n = 11, 12, 13 and 14 are conjectured values supported by numerical results (see Packomania in the links).
It can be shown that a(n) <= n for n >= 3. As n increases, terms of values other than 3 and 6 will eventually disappear. For example, the packing density of triangular packing of more than 121 circles inside an equilateral triangle, or hexagonal packing of more than 552 circles inside a regular hexagon, is higher than that of square packing inside a square. Thus, for n > 121, the sequence does not have any terms with a(n) = 4.
Conjecture: As n tends to infinity, a(n) takes the value of 3 or 6 and the packing density approaches sqrt(3)*Pi/6.

Examples

			a(1) = 0. The maximum packing density for packing 1 circle in regular m-gon is (Pi/m)*cot(Pi/m), which is an increasing function of m. Highest packing density of 1 is achieved as m tends to infinity and the regular n-gon becomes a circle.
a(2) = 4. The maximum packing density for packing 2 circles in regular polygon with odd number of edges m >= 3 is 4*Pi/(m*sin(2*Pi/m))/(sec(Pi/(2*m))+sec(Pi/m))^2, which is smaller than the packing density in regular polygon with even number of edges m >= 4, 4*Pi/(m*sin(2*Pi/m))/(1+sec(Pi/m))^2, which is a decreasing function of m with a maximum of Pi/(3+2*sqrt(2)) at m = 4.
Symmetry type (S) of the n-circle configuration achieving the highest packing density and the corresponding number of edges (N) of the regular polygon and packing density for n up to 16 are listed below.
  n        S     N    Packing density
------  ------  ---   ---------------------------------------------------------
1       O(2)    oo                                                     1
2       D_{4}    4    Pi/(3+2*sqrt(2))                               = 0.53901+
3       D_{6}    3    (Pi/2)/(1+2/sqrt(3))                           = 0.72900+
4,9,16  D_{8}    4    Pi/4                                           = 0.78539+
5       D_{10}   5    (Pi/2)/(1+4/sqrt(10+2*sqrt(5)))                = 0.76569+
6       D_{6}    3    6*Pi/(12+7*sqrt(3))                            = 0.78134+
7       D_{12}   6    7*Pi/(12+8*sqrt(3))                            = 0.85051+
8       D_{14}   7    (4*Pi/7)/(1+1/sin(2*Pi/7))                     = 0.78769+
10      D_{6}    3    (5*Pi/3)/(3+2*sqrt(3))                         = 0.81001+
11      D_{2}    9    (11*Pi/18)/(1+csc(2*Pi/9))                     = 0.75120+
12      D_{6}    6    6*Pi/(12+7*sqrt(3))                            = 0.78134+
13      D_{2}   10    (13*Pi/20)/(1+sqrt(50+10*sqrt(5))/5)           = 0.75594+
14      D_{4}    6    (49*Pi/2)/(21+20*sqrt(3)+6*sqrt(7)+6*sqrt(21)) = 0.77737+
15      D_{6}    3    15*Pi/(24+19*sqrt(3))                          = 0.82805+
		

Crossrefs

Showing 1-5 of 5 results.