As Alan W noted, the problem was poorly stated in my initial email. Here is a simplified version: Let S(n, k) = [n/2^k + 1/2] What are f(n) = SUM(k = 1 to inf; S(n, k)) g(n) = SUM(k = 1 to inf; 2^k * S(n, k)^2) Paul H essentially got the answer: f(n) = n g(n) = n^2 + n The fun part is the proof.