Keith, Bearing in mind your remark about tuits, I created A327448 and A327449 for cubes and primes. If you have others (squares?), please post them here - or email them to me. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Tue, Sep 17, 2019 at 9:30 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
"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
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