some more (minor) results for the 3-parameter case. recall that my notation is: T(n_1, n_2, ... , n_k) is the largest c such that there is a partition of 1 with all parts >= c which is a simultaneous refinement of 1/n_i + 1/n_i + ... + 1/n_i = 1 for each i . i noticed that there are examples of optimal partitions for T(m, n) which have some 1/n 's unsplit, and others split. unless the minimal part is 1/(2n) (or 1/n in the case that m divides n ), we have some freedom in choosing how to split the previously unsplit 1/n 's, and sometimes we can get optimal partitions for the 3-parameter case. for example: T(5, 8) = 1/20 , from [4 * 1/20] + 4 * [3/40 + 1/8] <---> 4 * [1/20 + 3/40] + 4 * [1/8] . by splitting two of the 1/8 's into 7/120 + 1/15 , we also get a refinement of 1/3 + 1/3 + 1/3 : [3 * 1/20 + 7/120 + 1/8] + [1/20 + 7/120 + 3 * 3/40] + [2 * 1/15 + 3/40 + 1/8] , which shows that T(3, 5, 8) = 1/20 . T(6, 8) = 1/24 , from 6 * [1/24 + 1/8] <---> 2 * [3 * 1/24] + 6 * [1/8] . by splitting the 1/8 's as 7/120 + 1/15 , we also get a refinement of 1/5 + 1/5 + 1/5 + 1/5 + 1/5 : 3 * [2 * 1/24 + 2 * 7/120] + 2 * [3 * 1/15] , which shows that T(5, 6, 8) = 1/24 . instead, we might split 3 of the 1/8 's as 1/21 + 13/168 , and the other 3 as 5/84 + 11/168 , to get a refinement of 7 * 1/7 : 3 * [1/24 + 1/21 + 13/168] + 3 * [1/24 + 5/84 + 11/168] + [3 * 1/21] , which shows that T(6, 7, 8) = 1/24 . alternatively, if we split each 1/8 as 1/18 + 5/72 , we get a refinement of 9 * 1/9 : 6 * [1/24 + 5/72] + 3 * [2 * 1/18] , which shows that T(6, 8, 9) = 1/24 .
Muffin (5,6,7) = muffin (2,3,5,6,7) = 1/21: 210/7 = [10.10.10], 2[11.19], [12.18], [13.17], 2[14.16] 210/6 = [10.12.13], 2[10.11.14], [17.18], 2[16.19] 210/5 = [16.16.10], [17.11.14], [18.10.14], [19.10.13], [19.11.12]
very nice! i noticed that there was a continuum of optimal partitions for T(5, 7) when i briefly tried to determine if some of the optimal partitions were unique, but did not make the connection to using this to try for optimal partitions for the 3-parameter case. this is better than what i did above, because the above is a special case of having continua of optimal partitions.
I believe this partition is unique up to piece size and multiplicity, and that muffin(4,5,6,7) < 1/21.
the latter statement would follow from the former, so i tried to prove the former. but it is not true, as i found several more optimal partitions for T(5, 6, 7): (1) [1/21 + 2/35 + 2/21] + [1/21 + 13/210 + 19/210] + [1/21 + 1/15 + 3/35] + [1/21 + 1/14 + 17/210] + [11/210 + 1/14 + 8/105] <---> [2 * 1/21 + 1/14] + [1/21 + 11/210 + 1/15] + [1/21 + 2/35 + 13/210] + [1/14 + 2/21] + [8/105 + 19/210] + [17/210 + 3/35] <---> [3 * 1/21] + [1/21 + 2/21] + [11/210 + 19/210] + [2/35 + 3/35] + [13/210 + 17/210] + [1/15 + 8/105] + [2 * 1/14] (2) [1/21 + 37/630 + 59/630] + [1/21 + 22/315 + 26/315] + [1/21 + 23/315 + 5/63] + [31/630 + 4/63 + 11/126] + [1/18 + 19/315 + 53/630] <---> [1/21 + 31/630 + 22/315] + [1/21 + 1/18 + 4/63] + [1/21 + 37/630 + 19/315] + [23/315 + 59/630] + [5/63 + 11/126] + [26/315 + 53/630] <---> [3 * 1/21] + [31/630 + 59/630] + [1/18 + 11/126] + [37/630 + 53/630] + [19/315 + 26/315] + [4/63 + 5/63] + [22/315 + 23/315] (3) [1/21 + 7/120 + 79/840] + [1/21 + 9/140 + 37/420] + [1/21 + 59/840 + 23/280] + [41/840 + 61/840 + 11/140] + [23/420 + 17/280 + 71/840] <---> [1/21 + 41/840 + 59/840] + [1/21 + 23/420 + 9/140] + [1/21 + 7/120 + 17/280] + [61/840 + 79/840] + [11/140 + 37/420] + [23/280 + 71/840] <---> [3 * 1/21] + [41/840 + 79/840] + [23/420 + 37/420] + [7/120 + 71/840] + [17/280 + 23/280] + [9/140 + 11/140] + [59/840 + 61/840] (4) [3 * 1/21 + 2/35] + [1/21 + 13/210 + 19/210] + [1/21 + 1/15 + 3/35] + [1/21 + 1/14 + 17/210] + [11/210 + 1/14 + 8/105] <---> 2 * [2 * 1/21 + 1/14] + [1/21 + 11/210 + 1/15] + [1/21 + 2/35 + 13/210] + [8/105 + 19/210] + [17/210 + 3/35] <---> 2 * [3 * 1/21] + [11/210 + 19/210] + [2/35 + 3/35] + [13/210 + 17/210] + [1/15 + 8/105] + [2 * 1/14] partition (4) is derived from (1) by splitting the 2/21 into 2 * 1/21 . i got these by classifying all optimal partitions for the (5, 7) case, and then examining those for special cases that are also refinements of 6 * 1/6 . for the (5, 7) case, i found 9 2-dimensional families, ( 6 parallelogram regions and 3 hexagonal regions), 9 1-dimensional families, and 3 isolated cases. that gives some idea of the complexity of the problem. i think that i found all solutions to the (5, 7) case and also to the (5, 6, 7) case, which would imply that indeed T(4, 5, 6, 7) < 1/21 . however, i am not overly confident that i didn't miss a sub-sub-sub-subcase somewhere! it is easy to specialize the aforementioned families to find a partition that establishes T(4, 5, 7) = 1/21 , for instance, [1/21 + 1/20 + 1/15 + 3/35] + [1/21 + 2 * 5/84 + 1/12] + [1/21 + 2/35 + 29/420 + 8/105] + [31/420 + 1/12 + 13/140] <---> [1/21 + 5/84 + 13/140] + [1/21 + 1/15 + 3/35] + [1/21 + 29/420 + 1/12] + [1/20 + 31/420 + 8/105] + [2/35 + 5/84 + 1/12] <---> [3 * 1/21] + [1/20 + 13/140] + [2/35 + 3/35] + 2 * [5/84 + 1/12] + [1/15 + 8/105] + [29/420 + 31/420] also, a partition that demonstrates T(4, 6, 7) = 1/21 is easy to find: [2 * 1/21 + 5/84 + 2/21] + [3 * 5/84 + 1/14] + [1/14 + 1/12 + 2/21] + [3 * 1/12] <---> 2 * [1/21 + 2 * 5/84] + 2 * [1/14 + 2/21] + 2 * [2 * 1/12] <---> 2 * [1/21 + 2/21] + 4 * [5/84 + 1/12] + [2 * 1/14] . mike