Let f(m, n) = 3m+n+(m+n)^{2}. Find ∑_{0}^{∞}∑_{0}^{∞} 2^{-f(m, n)}.

**Solution**

Answer: 4/3.

The obvious approach is to sum over n and then over m (or vice versa). But it is not obvious how to sum x^{N} where N runs through the squares. So we wonder just what values f(m, n) can take. A little experimentation suggests that it takes each even value just once.

In fact, it is clear that r(r+1) is always even and that the difference between r(r+1) and the next number in the sequence, (r+1)(r+2) is 2r+2, so any positive even number can be uniquely expressed as r(r+1) + 2s with 0 ≤ s ≤ r. But taking m = s, n = r-s, this is equivalent to the statement that any positive even number can be uniquely expressed as 3m+n+(m+n)^{2}.

Hence the sum is just 1 + 1/4 + 1/4^{2} + ... = 4/3.

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002