Very Nice BruceZ.


o(2) = 4

t(2)= 1

o(n) = 2[o(n-1)-t(n-1)]

t(n) = [o(n-1)-2*t(n-1)]/2


If we use standard techniques for recurrence relations, then we can solve those equations.

(I will use the Mathematica algebra package because I am lazy.) The result is


t(n) = (5 - Sqrt[5])/10* ((1 + Sqrt[5])/2)^n + (5 + Sqrt[5])/10* ((1 - Sqrt[5])/2)^n


which is Fibonacci as pointed out by lorinda.


Two questions for bruceZ:


1) How did you figure out t(n) = [o(n-1)-2*t(n-1)]/2?

2) How did you format the table? The last time that I posted actively in this forum, I used to be able to include html key words in these posts. Now those key words don't seem to have an effect.