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A244093

Rounded down ratio of area of a unit circle and a circle inscribed in any of the n triangles composing a regular n-gon which is circumscribed by a unit circle.

18, 11, 11, 12, 13, 15, 17, 19, 22, 25, 28, 31, 35, 39, 42, 47, 51, 56, 60, 65, 70, 76, 81, 87, 93, 99, 106, 112, 119, 126, 133, 141, 148, 156, 164, 173, 181, 190, 198, 207, 217, 226, 236, 246, 256, 266, 276, 287, 298, 309, 320, 332, 343, 355, 367, 380, 392, 405, 418, 431, 444

(list; graph; refs; listen; history; text; internal format)

OFFSET

3,1

COMMENTS

The minimum ratio occurs at n = 5.

LINKS

Table of n, a(n) for n=3..63.

Kival Ngaokrajang, Illustration of initial terms

FORMULA

a(n) = floor(Pi/area(n)) where area = Pi*r(n)^2, r(n) = (s(n)/2)*sqrt((2 - s(n))/(2 + s(n))), with s(n) = 2*sin(Pi/n) which is the side length (length unit 1) of the regular n gon. [rewritten by Wolfdieter Lang, Jun 30 2014 and Jul 02 2014]

a(n) = floor(1/r(n)^2) with r(n) = S(n)*(1 + C(n) - S(n))/(1 + C(n) + S(n)) with S(n) = s(n)/2 and C(n) = cos(Pi/n). 2*C(n) is the ratio of the length of the smallest diagonal and the side length s(n) in the regular n-gon. - Wolfdieter Lang, Jun 30 2014

PROG

(PARI)

{

for (n=3, 100,

c=2*sin(Pi/n);

s=(2+c)/2;

r=sqrt(((s-1)^2*(s-c))/s);

area=Pi*r^2;

a=floor(Pi/area);

print1(a, ", ")

)

}