%I #48 Oct 28 2019 20:41:25

%S 4,8,20,12,28,16,12,20,44,24,52,28,20,32,68,36,76,40,28,44,92,48,100,

%T 52,36,56,116,60,124,64,44,68,140,72,148,76,52,80,164,84,172,88,60,92,

%U 188,96,196,100,68,104,212,108,220,112,76,116,236,120,244,124,84

%N Denominator of time taken for a vertex of a rolling regular n-sided polygon to reach the ground.

%C 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).

%C Lim_{n->infinity} b(n) = 1/2 (can be easily proved).

%F a(n) = denominator((n - 1) / (2*n)) for even n; a(n) = denominator((2*n - 3) / (4*n)) for odd n.

%e For n = 3, a(3) = denominator of ((2*3-3)/4*n) = denominator of (3/12) = denominator of (1/4) = 4.

%e 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.

%t Array[Denominator[(2 (# - 1) - Mod[#, 2])/(4 #)] &, 61, 3] (* _Michael De Vlieger_, Oct 06 2019 *)

%o (PARI) a(n) = {denominator((2*(n-1) - n%2)/(4*n))} \\ _Andrew Howroyd_, Oct 06 2019

%Y Cf. A328185 (numerators).

%K nonn,frac

%O 3,1

%A _Luca Alexander_, Oct 06 2019