Michael, your partition for (23,13) and proof of optimality are both very impressive! The bound S(23,13) >= 21/52 is my "strong fill constraint", findable by a simple algorithm. For m/p with gcd(p,5) = 1 and 8/5 <= m/p <= 9/5, it says S(m,p) >= (2k+1)/(5k+2) at the least k where p divides 5k+2, achievable by a partition with relative integral sizes in 2k+1..3k+1. Your impressive example does even better than the strong fill constraint. When 5 divides p and 8/5 <= m/p <= 9/5 I don't know to do better than S(m,p) > 2/5 (my "weak fill constraint"), but your example suggests it might be possible in some cases.
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] .
okay, i can now prove that T(13, 23) = 53/2990 . this is the most involved case i've done, but it becomes much easier after finding the right way to organize things. [snip]