OFFSET
3,1
COMMENTS
Given an n-sided regular polygon "rolling" on a flat surface with constant angular velocity, a(n) is the denominator of the ratio: [("time" taken for any one vertex to move from highest point to lowest point) / ("time" taken for polygon to finish one complete turn)] := b(n).
Lim_{n->infinity} b(n) = 1/2 (can be easily proved).
FORMULA
a(n) = denominator((n - 1) / (2*n)) for even n; a(n) = denominator((2*n - 3) / (4*n)) for odd n.
EXAMPLE
For n = 3, a(3) = denominator of ((2*3-3)/4*n) = denominator of (3/12) = denominator of (1/4) = 4.
a(4) = 8 since it takes 3/8 of a full revolution of a square for a vertex to go from the highest point to the lowest point. When the vertex is at its highest position the square will be oriented at 45 degrees to the plane.
MATHEMATICA
Array[Denominator[(2 (# - 1) - Mod[#, 2])/(4 #)] &, 61, 3] (* Michael De Vlieger, Oct 06 2019 *)
PROG
(PARI) a(n) = {denominator((2*(n-1) - n%2)/(4*n))} \\ Andrew Howroyd, Oct 06 2019
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Luca Alexander, Oct 06 2019
STATUS
approved