On Thu, 5 Mar 2009, I wrote:
Okay, it turns out U(6, n) wasn't very hard, and I got around to going through it the other day.
I skipped over U(5, n), since there were several cases, but I went through them, and it's not so bad. In general, when considering U(m, n), you have to think about 2n mod m, but then sometimes you have to look at cases based on (roughly) the number of pieces an n splits into, modulo 2n mod m. Anyway, for n>=3m/2 (n>=8) (and n not a multiple of 5), the values for U(5, n) all achieve equality in the general upper bound U(m, n) <= min( n/([2n/m]+1), m - n/[2n/m] ), except for the case U(5, 11)=13/6, which I discussed earlier. Here are the optimal partitions for n at least 5 and not 7 or 11. I write them with mixed fractions, and I use "-+" to mean that the indicated number of pieces occur both with the "-" sign and with the "+" sign. Also, for a half an odd integer, I write m * a to mean m/2 * 1, m * (a-1/2). Then, whether a is integral or not, I'll write (m * a)b to mean b copies of m * a. Finally, I write e for either -1 or +1. These are all optimal for k>=1 with the exceptions of (5, 7) (for which you can do better) and (5, 11) (for which the construction doesn't work). U(5, 5k): 5 * 5k U(5, 10k+2e): [5/2 -+ 1/2k] * 4k, (5 * (3k+e)/2)4 U(5, 10k+3e): [5/2 -+ 1/(2k+e)] * (4k+2e), [5/2 -+ 1/(2k+e)(6k-1+e)] * (6k-1+e), 5/2 * 2 U(5, 15k+e): [5/2 -+ 3/(12k+2e)] * (12k+2e), [5/2 -+ 1/(12k+2e)k] * 2k, (5 * (k-e)/2)2 U(5, 15k+4e): [5/2 -+ 3/(12k+2e)] * (12k+2e), [5/2 -+ 1/(12k+2e)(k+e)] * (2k+2e), (5 * k/2)2 U(5, 15k+6e): [5/2 -+ 1/(4k+2e)] * (12k+6e), 5 * 3k As usual, one can check that these are valid co-refinements of 5 * n and n * 5, and the standard upper bound above shows that they achieve U. And a bit more checking shows that they are optimal partitions. Presumably, for any fixed value of m, one can proceed in this manner, but there will probably be more cases and more exceptions for small values of n. David P. Moulton