[math-fun] Variation on Dan's puzzle

Let f(n) be the number of prime factors of n, and let g(n) be the number of distinct prime factors of n, e.g. f(12)=f(2.2.3)=3, g(12)=2. Let F(n) = (-1)^f(n), G(n) = 2^g(n). Then: sum(F(d) G(n/d), d divides n) = 1. Or, the other way around, F(n) = 2^f(n), G(n) = (-1)^g(n), and the result is also true. Now let F(n) = (1/2)^f(n), G(n) = (1/2)^g(n), and again the same conclusion is true. In fact, let F(n) = a^f(n), G(n)=b^g(n), a+b=1, and the conclusion continues to hold. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com