23/13: pieces 21..31: 21.31*12, 22.30*6, 23.29*4, 26.26; 31.31.30*6, 21.21.21.29*4, 22.22.22.26*2, 23.23.23.23 S = 21/52, R_U = 31/21
ouch! i can't manage anything with this one yet ...
double ouch!! the combinatorics here seem to be a bit involved. however, i can improve this to T(13, 23) >= 53/2990 with the partition 2 * [3 * 53/2990 + 71/2990] + 2 * [2 * 53/2990 + 59/2990 + 1/46] + 2 * [2 * 107/5980 + 27/1495 + 3/130] + [2 * 27/1495 + 2 * 61/2990] + 4 * [38/1495 + 2 * 77/2990] + 2 * [2 * 153/5980 + 77/2990] <---> 10 * [53/2990 + 77/2990] + 4 * [107/5980 + 153/5980] + 4 * [27/1495 + 38/1495] + 2 * [59/2990 + 71/2990] + 2 * [61/2990 + 3/130] + [2 * 1/46] . no guarantee that this is best possible. i also briefly looked into the 3-parameter case. recall my notation is T(m, n, p) = maximum c for which there is a partition of 1 with all parts >= c , and which is a refinement of 1/m + 1/m + ... + 1/m = 1 , and of 1/n + 1/n + ... + 1/n = 1 and 1/p + 1/p + ... + 1/p = 1 . i think i checked all the possibilities to show that T(3, 4, 5) = 1/15 . the analysis is similar to the 2-parameter case to find the different types of splitting of individual fractions, but there are many different ways the parts can be redistributed in the other splittings. this was the bulk of the analysis. by way of contrast, it was much easier to prove that T(3, 5, 7) = 1/21 . partitions and proofs left to the interested reader. mike