A084619 Duplicate of A023393.
0, 1, 2, 7, 11, 19, 27, 38, 50, 64, 80, 98, 118
Offset: 1
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.
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.
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.
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
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+
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+
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