OFFSET
0,3
COMMENTS
See illustration in link for construction rule.
From Wolfdieter Lang, Apr 01 2014: (Start)
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)
LINKS
Kival Ngaokrajang, Illustration of initial terms
FORMULA
a(n) = 2*floor((1-r)/(2*r*cos(Pi/6))) + 1, where r = 1/2^n.
a(n) = 2*floor(sqrt((2^n-1)^2/3)) + 1. - Andrew Howroyd, May 27 2018
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, Mar 12 2014
STATUS
approved