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A239206
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a(n) is the total number of rows of circles of radius r packing into a circle of radius R, where r = R/2^n.
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3
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1, 1, 3, 9, 17, 35, 73, 147, 295, 591, 1181, 2363, 4729, 9459, 18917, 37837, 75673, 151347, 302697, 605395, 1210791, 2421581, 4843163, 9686329, 19372659, 38745319, 77490641, 154981281, 309962565, 619925129, 1239850261, 2479700523, 4959401047, 9918802097
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OFFSET
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0,3
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COMMENTS
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See illustration in link for construction rule.
Call in the above mentioned illustration (link) the horizontal layers with the y-axis (center as origin) intersecting a circle B-layers and the other ones A-layers. Consider first only the upper half, with the A-layer y=0 treated separately. The largest k value for B-layers with yB(k) = sqrt(3)*(2*k + 1), k >= 0, is kBmax(n) = floor((2^n-1-sqrt(3))/(2*sqrt(3))), n>=2. Similarly, the largest k value for A-layers with yA(k) = sqrt(3)*2*k, k >= 0, is kAmax(n) = floor(sqrt((2^n-1)^2-1)/(2*sqrt(3))), n >= 1. For n=0 one has the large disk. If the top level is of B-type then 2*kBmax(n) + 1 > 2*kAmax(n). In this case the total number of layers (A or B) is 4*kBmax(n) + 3, n >= 2. In the other case (top level is of the A-type) it is 1 + 4*kAmax(n), n >= 1. With a(0) = 1 Kival Ngaokrajang's formula given below, which is a(n) = 2*floor((2^n-1)/sqrt(3)) + 1, seems to fit these numbers. (End)
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LINKS
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FORMULA
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a(n) = 2*floor((1-r)/(2*r*cos(Pi/6))) + 1, where r = 1/2^n.
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MATHEMATICA
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rcr[n_]:=Module[{r=1/2^n, c=Cos[Pi/6]}, 2*Floor[(1-r)/(2*r*c)]+1]; Array[ rcr, 40, 0] (* Harvey P. Dale, Dec 23 2014 *)
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PROG
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(PARI) a(n) = 1 + 2*sqrtint((2^n-1)^2\3); \\ Andrew Howroyd, May 27 2018
(Small Basic)
For n = 0 To 50
r = 1/Math.Power(2, n)
a = 2*math.Floor((1-r)/(2*r*math.Cos(Math.Pi/6))) + 1
TextWindow.Write(a+", ")
EndFor
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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