OFFSET
0,1
COMMENTS
Since the perimeter equals n*tan(180ยบ/n), increasing n to greater values will yield a more accurate value of Pi.
Lim n -> inf., a(n+1)/a(n) = sqrt(10). This implies that a(n+2) ~ 10*a(n).
Lim n -> inf., a(2n) = 10^n*sqrt(Pi^3/3) and a(2n+1) = 10^n*sqrt(Pi^3/30).
Lim n -> inf., a(n)/A259441(n) = sqrt(2).
REFERENCES
William H. Beyer, Ed., CRC Standard Mathematical Tables, 27th Ed., IV - Geometry, Mensuration Formulas, p. 122, Boca Raton 1984.
Daniel Zwillinger, Editor-in-Chief, 31st Ed., CRC Standard Mathematical Tables and Formulae, 4.5.3 Geometry - Regular Polygons, p. 324, Boca Raton, 2003.
Jan Gullberg, Mathematics: From the Birth of Numbers, 13.3 Solving Triangles, p. 479, W. W. Norton & Co., NY, 1997.
Catherine A. Gorini, Ph.D., The Facts on File Geometry Handbook, Charts & Tables, p. 262, Checkmark Books, NY, 2005.
LINKS
mathematicsonline, How to Calculate Pi using Archimedes' Method
EXAMPLE
a(0) # 3 since the perimeter of the circumscribed triangle is sqrt(27) which equals approximately 5.196152... which exceeds Pi by more than 1;
a(0) = 4 since the perimeter of the circumscribed square is 4 and this is within 1 of the true value of Pi;
a(1) = 11 since the perimeter of the circumscribed 11-gon which equals approximately 3.229891... which is within 0.1 of the true value of Pi;
a(2) = 33 since the perimeter of the circumscribed 33-gon which equals approximately 3.151117... which is within 0.01 of the true value of Pi; etc.
MATHEMATICA
f[n_] := Block[{k = Floor[ Sqrt[ 10]*f[n - 1]] - 6}, While[Pi + 10^-n < k*Tan[Pi/k], k++]; k]; f[-1] = 3; Array[f, 28, 0]
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Robert G. Wilson v, Jun 27 2015
STATUS
approved