Let f(m, n) = 3m+n+(m+n)2. Find ∑0∞∑0∞ 2-f(m, n).
The obvious approach is to sum over n and then over m (or vice versa). But it is not obvious how to sum xN 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 + 1/42 + ... = 4/3.
21st Putnam 1960
© John Scholes
15 Feb 2002