"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
Interestingly, the solution for cubes, A113263, is phrased in OEIS as "a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can be partitioned into two sets of equal sums." I think I like that phrasing better than the "sum to zero" phrasing, as it leads to further questions, such as how many ways a set can be partitioned into *three* sets of equal sums.
Surprisingly, that isn't in OEIS, nor for primes, nor for squares. I'll add them when I get a tuit with fewer corners. Here's how many ways the first n cubes can be partitioned into three sets of equal sums, starting with n=23: 1,0,0,691,3416,0,233,1168,0,17714,30601,0,3087050,... Here's the unique first solution: 27 + 216 + 1000 + 2197 + 5832 + 6859 + 9261 = 1 + 64 + 343 + 512 + 1728 + 4096 + 8000 + 10648 = 8 + 125 + 729 + 1331 + 2744 + 3375 + 4913 + 12167 And here's the sequence for primes, starting with n=10: 3,0,0,0,0,0,423,0,2624,0,13474,0,611736,0,4169165,0,30926812,0,... (No, every other term isn't always zero.) And here's one of the first solutions for primes: 3 + 17 + 23 = 2 + 5 + 7 + 29 = 11 + 13 + 19