%I
%S 3,5,9,13,19,24,32,38,48,56,67,77,90,102,116,129,145,160,178,194,213,
%T 231,252,272,294,316,340,363,388,413,440,466,495,523,554,583,615,646,
%U 680,713,748,782,820,855,894,932,972,1011,1053,1094,1137,1180,1225
%N How small is the squeezed ngon? Let s0 be the side of a regular ngon and s1 the side of the maximal ngon which can be squeezed between the former and its circumcircle. The nth entry in the sequence is floor(s0/s1).
%C Closely related to K(n) = (2*n/Pi)*sin(Pi/n)/(1cos(Pi/n)) as derived from the ngon with same circumference as the circle squeezed between the large ngon and its circumcircle.
%H Robert Israel, <a href="/A065802/b065802.txt">Table of n, a(n) for n = 3..10000</a>
%H Bill Taylor, Ignacio Larrosa CaĆ±estro, Rainer Rosenthal, <a href="https://groups.google.com/d/msg/sci.math/BoYg5fBSvs/IoMzfdUGC3EJ">Little Geometry Problem</a>, thread in newsgroup sci.math, Oct 31  Nov 06, 2001.
%F For n=odd: a(n) = floor((1+cos(Pi/n))/(1cos(Pi/n))) For n=even: a(n) = floor( 2*(2/(tan(Pi/n))^2) + 1 )
%F 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
%e 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.
%p f:= proc(n) if n::odd then floor((1+cos(Pi/n))/(1cos(Pi/n))) else floor(2*(2/(tan(Pi/n))^2) + 1) fi end proc:
%p map(f, [$3..100]); # _Robert Israel_, Oct 24 2017
%t 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} ]
%Y Cf. A055684.
%K easy,nice,nonn
%O 3,1
%A _Rainer Rosenthal_, Dec 05 2001
%E More terms from _Robert G. Wilson v_, Dec 06 2001
