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Could muffin(4k,5k,6k,7k) = 1/21 for some k?

surely you mean 1/(21k) above. and yes, n = 2 works: [ ... ]

k = 3 also works: [2 * 1/63 + 1/45 + 37/1260] + [2 * 1/63 + 8/315 + 11/420] + [1/63 + 2 * 17/840 + 17/630] + [1/63 + 23/1260 + 13/630 + 1/35] + 2 * [2 * 1/63 + 61/2520 + 23/840] + [41/2520 + 11/630 + 59/2520 + 11/420] + [2/105 + 7/360 + 3/140 + 59/2520] + [3/140 + 19/630 + 2/63] + 2 * [1/42 + 71/2520 + 79/2520] + [41/2520 + 23/1260 + 7/360 + 37/1260] <---> [1/63 + 2/105 + 2/63] + [1/63 + 13/630 + 19/630] + [1/63 + 1/45 + 1/35] + [1/63 + 1/42 + 17/630] + [11/630 + 1/42 + 8/315] + 2 * [1/63 + 7/360 + 79/2520] + 2 * [1/63 + 3/140 + 37/1260] + 2 * [1/63 + 59/2520 + 23/840] + 2 * [41/2520 + 61/2520 + 11/420] + 2 * [23/1260 + 17/840 + 71/2520] <---> [2 * 1/63 + 1/42] + [1/63 + 11/630 + 1/45] + [1/63 + 2/105 + 13/630] + [1/42 + 2/63] + [8/315 + 19/630] + [17/630 + 1/35] + 2 * [1/63 + 41/2520 + 59/2520] + 2 * [1/63 + 23/1260 + 3/140] + 2 * [1/63 + 7/360 + 17/840] + 2 * [61/2520 + 79/2520] + 2 * [11/420 + 37/1260] + 2 * [23/840 + 71/2520] <---> 3 * [3 * 1/63] + [1/63 + 2/63] + [11/630 + 19/630] + [2/105 + 1/35] + [13/630 + 17/630] + [1/45 + 8/315] + [2 * 1/42] + 2 * [41/2520 + 79/2520] + 2 * [23/1260 + 37/1260] + 2 * [7/360 + 71/2520] + 2 * [17/840 + 23/840] + 2 * [3/140 + 11/420] + 2 * [59/2520 + 61/2520] i got this in much the same way: combining 3 optimal partitions for the (5, 6, 7) case and rescaling. then it was a simple matter to arrange the parts to get a refinement of 12 * 1/12 . moreover, by combining the k = 2 and k = 3 cases, we get partitions that prove T(4k, 5k, 6k, 7k) = 1/(21k) for all k > 1 . my earlier search for optimal partitions for the (5, 6, 7) case shows, modulo possibly missing some sub-sub-subcases, that T(4, 5, 6, 7) < 1/21 , as scott surmised. i suppose now i should scrutinize the details of this search to make sure i didn't miss anything. mike