

A065802


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).


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
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OFFSET

3,1


COMMENTS

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.


REFERENCES

Bill Taylor, "Little Geometry problem", Newsgroup sci.math, 31Oct2001


LINKS

Robert Israel, Table of n, a(n) for n = 3..10000


FORMULA

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 )
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))/(1cos(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



