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 1-9 of 9 results.

A084619 Duplicate of A023393.

Original entry on oeis.org

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

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Author

Keywords

A084617 Maximum number of circles with diameter 1 that can be packed in a square of side length n.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 68, 86, 106, 128, 152, 181, 216, 247, 280
Offset: 1

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Author

Hugo Pfoertner, Jun 01 2003

Keywords

Comments

For n>5 the sequence terms are only conjectures. For more information see comment given in A084616.

References

Crossrefs

Extensions

More terms from Sergio Pimentel, Aug 08 2006

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.

A084644 Best packings of m>1 equal circles into a larger circle setting a new density record.

Original entry on oeis.org

2, 3, 4, 7, 19, 37, 55, 85, 121, 147, 148, 150, 151, 187, 188, 190, 191, 192, 193, 198, 199, 235, 241, 264, 267, 269, 270, 291, 292, 293, 294, 295, 343, 346, 347, 348, 349, 408, 409, 412, 414, 415, 417, 418, 419, 420, 421, 481, 499, 564, 565, 649, 689, 690, 721
Offset: 1

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Author

Hugo Pfoertner, Jun 01 2003

Keywords

Comments

Sequence terms for n>5 are only conjectures. The arrangement of 37 circles consists of one central circle surrounded by 3 rings of 6,12 and 18 circles. For n=7, 49 of the 55 circles are arranged in a rigid hexagonal lattice with 6 "rattlers" inserted in gaps at the circumference.

Examples

			a(4)=7 because the density 0.7777.. of the best packing of 7 circles (1 central circle surrounded by 6 neighbors) exceeds the density 0.68629.. of the packing of 4 circles arranged in a square.
		

References

  • List of references given by E. Specht; see corresponding link.

Crossrefs

Cf. A051657 (density records for circles packed into square), A084618, A023393.

Extensions

More terms from Robert G. Wilson v, Nov 07 2012

A084828 Maximum number of spheres of radius one that can be packed in a sphere of radius n.

Original entry on oeis.org

1, 2, 13, 32, 68
Offset: 1

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Author

Hugo Pfoertner, Jun 12 2003

Keywords

Comments

a(4) and a(5) are experimental values. Although A121346(5) claims a lower bound of a(5)=68, it is conjectured from extensive numerical search that this value is unachievable and therefore a(5)=67.
The conjecture a(5)=67 was proved wrong by Yu Liang, who found an arrangement of 68 spheres of radius 1 fitting into a sphere of radius 5.
Lower bounds for the next terms are a(6)>=122 and a(7)>=198. See E. Specht's webpage for latest data. - Hugo Pfoertner, Jan 22 2024

Crossrefs

Cf. A121346 (conjectured lower bounds), A084827, A084829, A084825.
Cf. A023393 (2D).

Extensions

Comment and links edited, a(5) from Hugo Pfoertner, Jun 23 2011
a(5) corrected, based on private communication from Yu Liang, by Hugo Pfoertner, Aug 24 2011

A253570 Maximum number of circles of radius 1 that can be packed into a regular n-gon with side length 2 (conjectured).

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 5, 7, 8, 9
Offset: 3

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Author

Felix Fröhlich, Jan 03 2015

Keywords

Comments

The values were obtained by constructing the circle arrangements in a vector graphics program and have not been proved to be correct.
From David Consiglio, Jr., Jan 09 2015: (Start)
As n increases, the n-gon more and more closely approximates a circle. As a result, the lower bound (which is highly likely to be the correct term for larger and larger n) is the number of circles that can be packed into an inscribed circle, the radius of which is given by the expression cot(Pi/n). Look up this radius in column 3 at www.packomania.com to find the lower bound of a(n).
A rough upper bound would be the closest packing of circles into the area of the n-gon (formula below). A better upper bound is likely possible.
See file for lower and upper bounds through a(20). The lower bounds have been proved for a(3) through a(13).
(End)

Crossrefs

Formula

Upper bound = floor(n/(2*sqrt(3)*tan(Pi/n))).

A201993 Conjectured lower bound for the number of circles of radius 1 that can be packed into a circle of radius n.

Original entry on oeis.org

1, 2, 6, 11, 18, 26, 37, 49, 63, 79, 97, 116, 138, 161, 186, 213, 241, 272, 304, 338, 374, 412, 451, 492, 535, 580, 627, 676, 726, 778, 832, 888, 946, 1005, 1066, 1130, 1194, 1261, 1330, 1400, 1472, 1546, 1622, 1699, 1779, 1860, 1943, 2028, 2115, 2203, 2293, 2385
Offset: 1

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Author

Hugo Pfoertner, Dec 07 2011

Keywords

Comments

Bound provided by David W. Cantrell in December 2008. It is conjectured that it is possible to find packings such that A023393(n)>=a(n) for all n. Currently (December 2011) the smallest number of circles, for which the bound is not achieved, is 507.

Crossrefs

Cf. A023393 (best known packings).

Programs

  • PARI
    for(k=2,53,my(rho=Pi/(2*sqrt(3)),N(R)=rho*R*(R-2)+R/2+1);print1(ceil(N(k-1)),", ")) \\ Hugo Pfoertner, Aug 02 2019

Formula

a(n) = Smallest k, such that 1 + (sqrt((4*Rho-1)^2 + 16*Rho*(k-1)) - 1) / (4*Rho) >=n with Rho = Pi/(2*sqrt(3)).

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

Views

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 1-9 of 9 results.