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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. the partition above establishes the lower bound. now suppose we have a partition with all parts >= 53/2990 . if any 1/23 is split into 3 or more parts, then the smallest is <= 1/69 , which is too small. since at least one 1/23 is split, there is no loss in generality in assuming that they are all split (by halving them, if necessary). thus there are 46 parts in total. if some 1/13 is split into 5 or more parts, then the smallest is <= 1/65 , which is too small. if some 1/13 is split into 2 parts, then the larger is >= 1/26 , so its "complement" in a 1/23 is <= 3/598 , which is too small. thus each 1/13 is split into either 3 or 4 parts. we now see that 7 of the 1/13 's are split into 4 parts, and the other 6 are split into 3 parts. we now consider how the parts pair up to form 1/23 's. of the 28 parts in the former 7 1/13 's, at most 18 can pair with a part in the latter 1/13 's, so there must be at least 5 pairs among these 28 parts. suppose that a subset of A of the former 1/13 's contains B pairs. the 4A parts sum to A/13 , and they contain 2B parts that sum to B/23 , so the other 4A - 2B parts sum to A/13 - B/23 = (23A - 13B)/299 . since each part is >= 53/2990 , we have (23A - 13B)/299 >= (53/2990)(4A - 2B) , which is equivalent to the inequality B <= 3A/4 . in particular, for A = 7 , which means all of the former 1/13 's, there are at most 5 pairs, so there must be exactly 5 pairs. now we claim that we can split these 7 1/13 's into two subsets so that none of the 5 pairs are separated. indeed, consider the graph on 7 vertices, corresponding to the 1/13 's, with an edge between two vertices if they contain parts that are paired with each other. since there are (at most) 5 edges, the graph is not connected. this proves the claim. now we see for various values of A , how many pairs a set of A 1/13 's can contain: A max. number of pairs = floor(3A/4) 1 0 2 1 3 2 4 3 5 3 6 4 we see that the only way to have two disjoint subsets that comprise all 7 1/13 's, and contain a total of 5 pairs, is one subset of 3 1/13 's having 2 pairs, and the other subset being the remaining 4 1/13 's , containing 3 pairs. finally, these 4 1/13 's have 16 parts summing to 4/13 , with the 6 parts in the 3 pairs summing to 3/23 . the remaining 10 parts sum to 4/13 - 3/23 = 53/299 , so the smallest of these is <= 53/2990 . this proves that T(13, 23) = 53/2990 . by way of contrast, it is quite easy to show that T(11, 19) = 9/418 . (this is another of scott's cases; i'm not sure if he claimed this was the optimal value, or just a lower bound.) if some 1/19 is split into 3 or more parts, the smallest is <= 1/57 , which is too small. some 1/19 must be split, so we may assume that all split. therefore, each 1/19 splits into exactly 2 parts, so there are 38 parts in total. is some 1/11 splits into 5 or more parts, the smallest is <= 1/55 , which is too small. if some 1/11 splits into 2 parts, the largest is >= 1/22 , so its complement in a 1/19 is <= 3/418 , which is too small. thus each 1/11 splits into either 3 or 4 parts, and there are 5 that split into 4 parts, and 6 that split into 3 parts. of the 20 parts in these former 1/11 's, at least 2 must pair up in the same 1/19 . the larger of these is >= 1/38 , so the other 3 parts in the same 1/11 sum to at most 1/11 - 1/38 = 27/418 . therefore the smallest of these 3 is <= 9/418 . this shows that T(11, 19) <= 9/418 . can we attain this value? well, the argument shows how to start. let me illustrate the parts by this array: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * where the columns correspond to the parts in the same 1/11 . now label the two parts in the left hand side that are "complements". they must be equal, and for each, the other three parts in the same 1/11 must be equal. thus we have a a * * * * * * * * * b b * * * * * * * * * b b * * * * * * * * * b b * * * and the complements of the b 's must occur on the right hand side. now let's make the remaining 12 parts on the left hand side equal; we get a a c c c * * * * * * b b c c c * * * * * * b b c c c * * * * * * b b c c c which forces the "complements" of the c 's to occur on the right hand side. there is still the task to organize the parts on the right hand side so that the columns each add to 1/11 . however, this is trivial! those parts come in half-dozens, 6 complements of b , call them d , and 12 complements of c , call them e , so put one d and two e 's in each column: a a c c c d d d d d d b b c c c e e e e e e b b c c c e e e e e e b b c c c where a = 1/38 , b = (1/11 - a)/3 = 9/418 , c = (1/11)/4 = 1/44 , d = 1/19 - b = 13/418 and e = 1/19 - c = 25/836 . (this seems to be different from scott's partition.) note also that all the parts are large enough; if not, the upper bound argument wouldn't be strong enough, but this would show exactly where to look for a better bound. mike