A337019
Maximum number of circles with unit area that can be packed into an equilateral triangle with an area of n.
Original entry on oeis.org
0, 1, 1, 1, 3, 3, 4, 6, 6, 6, 7, 8, 10, 10, 10, 11, 12, 13, 15
Offset: 1
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
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+
Cf.
A023393,
A051657,
A084616,
A084617,
A084618,
A084644,
A133587,
A227405,
A247397,
A253570,
A257594,
A269110,
A308578,
A337019,
A337020,
A343005.
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
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+
Cf.
A023393,
A051657,
A084616,
A084617,
A084618,
A084644,
A133587,
A227405,
A247397,
A253570,
A257594,
A269110,
A308578,
A337019,
A337020,
A343005,
A343262.
Showing 1-3 of 3 results.
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