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.
%I A239206 #21 May 27 2018 19:35:33
%S A239206 1,1,3,9,17,35,73,147,295,591,1181,2363,4729,9459,18917,37837,75673,
%T A239206 151347,302697,605395,1210791,2421581,4843163,9686329,19372659,
%U A239206 38745319,77490641,154981281,309962565,619925129,1239850261,2479700523,4959401047,9918802097
%N A239206 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.
%C A239206 See illustration in link for construction rule.
%C A239206 From _Wolfdieter Lang_, Apr 01 2014: (Start)
%C A239206 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)
%H A239206 Kival Ngaokrajang, <a href="/A239206/a239206.pdf">Illustration of initial terms</a>
%F A239206 a(n) = 2*floor((1-r)/(2*r*cos(Pi/6))) + 1, where r = 1/2^n.
%F A239206 a(n) = 2*floor(sqrt((2^n-1)^2/3)) + 1. - _Andrew Howroyd_, May 27 2018
%t A239206 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 *)
%o A239206 (PARI) a(n) = 1 + 2*sqrtint((2^n-1)^2\3); \\ _Andrew Howroyd_, May 27 2018
%o A239206 (Small Basic)
%o A239206 For n = 0 To 50
%o A239206 r = 1/Math.Power(2,n)
%o A239206 a = 2*math.Floor((1-r)/(2*r*math.Cos(Math.Pi/6))) + 1
%o A239206 TextWindow.Write(a+", ")
%o A239206 EndFor
%Y A239206 Cf. A239073, A239074.
%K A239206 nonn
%O A239206 0,3
%A A239206 _Kival Ngaokrajang_, Mar 12 2014