Re: [math-fun] Ulam(1,2)

Here's another idea to consider: Each Ulam number u(n), n >= 2 (zero index) is a unique sum a(n) + b(n) of distinct earlier Ulam numbers a(n) < b(n). Tabulate a(n) for 2 <= n <= 2000. There are only 47 distinct values of a(n). It seems like certain Ulams, starting with 2, 3, 47, 69, etc, are much more likely to be involved in the sums for subsequent Ulams.