On Fri, 16 Jan 2009, Michael Reid wrote:
i can now determine T(4, n) in general....
T(1, n) and T(2, n) are very easy to determine in general. it looks like there is probably also pattern in T(3, n) . T(6, n) might also be possible to determine in general.
Okay, it turns out U(6, n) wasn't very hard, and I got around to going through it the other day. We can assume n is relatively prime to 6, since otherwise you can multiply up a solution for U(2, n/3) or U(3, n/2), and it's still optimal. We already have optimal partitions for U(5, 6) and U(6, 7). Here are the rest; I decided to write mixed fractions, since that's how I think of them, and then I can use "-+" to mean take the parts using the "-" and also take the parts using the "+" (so there are five part sizes for each partition). Here they are U(6, 6k-1): [3 -+ 2/(2k-1)] * (4k-2), [3 -+ 1/k(2k-1)] * 2k, 3 * 2 (k >= 2) U(6, 6k+1): [3 -+ 2/(2k+1)] * (4k+2), [3 -+ 1/(k-1)(2k+1)] * (2k-2), 3 * 2 (k >= 2) One can check that they are valid co-refinements, and the standard upper bound shows that they achieve U. And a bit more checking shows that they are optimal partitions. David P. Moulton