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 A065802 How small is the squeezed n-gon? Let s0 be the side of a regular n-gon and s1 the side of the maximal n-gon which can be squeezed between the former and its circumcircle. The n-th entry in the sequence is floor(s0/s1). 1
 3, 5, 9, 13, 19, 24, 32, 38, 48, 56, 67, 77, 90, 102, 116, 129, 145, 160, 178, 194, 213, 231, 252, 272, 294, 316, 340, 363, 388, 413, 440, 466, 495, 523, 554, 583, 615, 646, 680, 713, 748, 782, 820, 855, 894, 932, 972, 1011, 1053, 1094, 1137, 1180, 1225 (list; graph; refs; listen; history; text; internal format)
 OFFSET 3,1 COMMENTS Closely related to K(n) = (2*n/Pi)*sin(Pi/n)/(1-cos(Pi/n)) as derived from the n-gon with same circumference as the circle squeezed between the large n-gon and its circumcircle. REFERENCES Bill Taylor, "Little Geometry problem", Newsgroup sci.math, 31-Oct-2001 LINKS Robert Israel, Table of n, a(n) for n = 3..10000 FORMULA For n=odd: a(n) = floor((1+cos(Pi/n))/(1-cos(Pi/n))) For n=even: a(n) = floor( 2*(2/(tan(Pi/n))^2) + 1 ) a(n) = floor(4*n^2/Pi^2) - b(n) where b(n) is in {0,1,2}; 0 occurs only for odd n, while 2 occurs only for even n. - Robert Israel, Oct 24 2017 EXAMPLE a(3) = 3 as can be seen in Christmas stars: cos(Pi/3)=1/2, thus a(3) = floor((3/2)/(1/2)) = 3. a(4) = 5 as proposed by Bill Taylor in sci.math: tan(Pi/4)=1, thus a(4) = floor(2*(2/1^2) + 1) = 5. MAPLE f:= proc(n) if n::odd then floor((1+cos(Pi/n))/(1-cos(Pi/n))) else floor(2*(2/(tan(Pi/n))^2) + 1) fi end proc: map(f, [\$3..100]); # Robert Israel, Oct 24 2017 MATHEMATICA f[n_] := If[ OddQ[n], Floor[(1 + Cos[Pi/n]) / (1 - Cos[Pi/n])], Floor[4/(Tan[Pi/n])^2 + 1] ]; Table[ f[n], {n, 3, 60} ] CROSSREFS Sequence in context: A212530 A004132 A207187 * A245302 A118028 A209974 Adjacent sequences:  A065799 A065800 A065801 * A065803 A065804 A065805 KEYWORD easy,nice,nonn AUTHOR Rainer Rosenthal, Dec 05 2001 EXTENSIONS More terms from Robert G. Wilson v, Dec 06 2001 STATUS approved

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Last modified December 6 14:15 EST 2019. Contains 329806 sequences. (Running on oeis4.)