17 Jan
2009
17 Jan
'09
7:04 a.m.
T(1, n) and T(2, n) are very easy to determine in general. it looks like there is probably also pattern in T(3, n) .
apparently U(3,3n) = 3 and U(3,3n +/- 1) = (3n-1)/2n. this translates to T(3,3n) = 1/3n, T(3,3n-1) = 1/6n, and T(3,3n+1) = (3n-1)/6n(n+1). got a proof mike? :) erich