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 1-8 of 8 results.

A023393 Maximal number of circles of radius 1 that can be packed in a circle of radius n.

Original entry on oeis.org

0, 1, 2, 7, 11, 19, 27, 38, 50, 64, 80, 98, 118
Offset: 0

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Author

Keywords

Comments

The terms for n>5 are only conjectures supported by extensive computations.

References

  • R. L. Graham and B. D. Lubachevsky, Dense packings of 3k(k+1)+1 equal disks in a circle for k = 1, 2, 3, 4 and 5, Proc. First Int. Conf. "Computing and Combinatorics" COCOON'95, Springer Lecture Notes in Computer Science 959 (1996), 303-312.
  • For list of references given by E. Specht, see the Specht link.

Crossrefs

Cf. A201993 (conjectured lower bounds for a(n)).

Extensions

Terms for n>5 from Hugo Pfoertner, Jun 01 2003
Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Feb 18 2004, writes to suggest that the sequence probably continues 138, 161, 187, 213, 242, 272, 304, 337, 373, 413, 451, 495, ...
Edited by N. J. A. Sloane at the suggestion of David W. Wilson, Sep 22 2007
Offset corrected by Jon E. Schoenfield, Oct 12 2008

A084616 Maximum number of circles of diameter 1 that can be packed in a square of area n (i.e., with side length n^(1/2)).

Original entry on oeis.org

1, 1, 2, 4, 4, 5, 5, 6, 9, 9, 9, 10, 12, 13, 14, 16, 16, 16, 18, 19, 20, 21, 22, 23, 25, 25, 26, 27, 28, 30, 30, 31, 33, 33, 34, 36, 36, 39, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 52, 52, 53, 53, 55, 56, 57, 58, 59, 59, 61, 62, 63, 65, 68, 68, 68, 69, 69, 70, 72, 73, 74, 74
Offset: 1

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Author

Hugo Pfoertner, Jun 01 2003

Keywords

Comments

Most sequence terms beyond n=20 are only conjectures supported by comprehensive numerical results. No proof is available for the following observations: n=30 is the first case where a square of area < n (29.74921576) is sufficient to cover n circles. The first case where more than n circles can be covered occurs for n=38. The required area to cover 39 circles is 37.76050335. n=59 is the last case where a square of area n does not suffice to cover n+1 circles (60 circles require square area 59.11626524).

Examples

			a(2)=1 because a square of side length sqrt(2)=1.414... is not large enough to cover two circles of diameter 1 (the required side length would be 1+sqrt(2)/2=1.707...).
a(38)=39 because 39 circles fit into a square of area 38.
		

Crossrefs

A084825 Maximum number of spheres of diameter one that can be packed in a cube of edge length n.

Original entry on oeis.org

1, 8, 27, 66
Offset: 1

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Author

Hugo Pfoertner, Jun 12 2003

Keywords

Comments

From an extrapolation of Dave Boll's numerical results a(4)~=66 and a(5)~=141 are estimated values for the next terms.
However, E. Specht's data suggest a(5)=135. - Hugo Pfoertner, Jul 08 2025

Examples

			a(3)=27 because there is no known better arrangement than the 3*3*3 cubic one that would allow packing more than 27 spheres into a cube of edge length 3.
		

Crossrefs

Extensions

a(4) from Hugo Pfoertner, May 21 2011

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

A308578 Maximum number of non-overlapping circles of radius 1/n that can be placed inside a unit square.

Original entry on oeis.org

0, 1, 1, 4, 5, 9, 10, 16, 20, 25, 30, 36
Offset: 1

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Author

Ethan D. Kidd, Jun 08 2019

Keywords

Comments

Alternatively described as the maximum number of circles of unit radius that can be placed inside a square of side length n.
It appears that the terms a(2) to a(8) are equal to the related terms of A189889.

Examples

			a(6)=9 because 9 circles of radius 1/6 can be placed in a 3 X 3 regular grid inside a unit square.
		

Crossrefs

Formula

a(2k) = A084617(k). - Jon E. Schoenfield, Jun 09 2019

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

A376215 Maximum number of non-overlapping discs with a diameter of 1 that can fit within an n×n square, using hexagonal packing or square packing, whichever is more efficient.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 68, 85, 105, 126, 150, 175, 216, 247, 279, 314, 350, 389, 429, 492, 538, 585, 635, 686, 740, 822, 880, 941, 1003, 1068, 1134, 1203, 1307, 1380, 1456, 1533, 1613, 1694, 1817, 1904, 1992, 2083, 2175, 2270, 2366, 2511, 2613, 2716, 2822, 2929, 3039
Offset: 1

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Author

Maurice Clerc, Sep 15 2024

Keywords

Comments

Maximum error on the first 99 terms = 3.3%. (See the Specht's site).
The first row of discs is placed on the base of the square. Then hexagonal packing is applied, row by row, as much as possible. We find a number N of discs. If n^2>N then the whole packing is replaced by the square one. Example: n=3 => N=8 but since n^2=9 we can indeed place 9 discs.
Note: useful only for n=2, 3, 4, 5, 6, 7.

Crossrefs

Cf. A084617.

Programs

  • Python
    from math import isqrt
    def A376215(n): return max(n**2,n*(m:=1+isqrt(((n-1)**2<<2)//3))-(m>>1)) # Chai Wah Wu, Nov 06 2024

Formula

a(n) = max(n^2,n*floor(1+2*(n-1)/sqrt(3)) - floor(floor(1+2*(n-1)/sqrt(3))/2));

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 1-8 of 8 results.