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.

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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

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