Could muffin(4k,5k,6k,7k) = 1/21 for some k?
surely you mean 1/(21k) above. and yes, n = 2 works: [2 * 1/42 + 1/30 + 37/840] + [2 * 1/42 + 4/105 + 11/280] + [1/42 + 41/1680 + 61/1680 + 17/420] + [1/42 + 23/840 + 13/420 + 3/70] + [1/42 + 17/560 + 59/1680 + 1/28] + [11/420 + 1/35 + 7/240 + 23/560] + [9/280 + 19/420 + 1/21] + [1/28 + 71/1680 + 79/1680] <---> [1/42 + 1/35 + 1/21] + [1/42 + 13/420 + 19/420] + [1/42 + 1/30 + 3/70] + [1/42 + 1/28 + 17/420] + [11/420 + 1/28 + 4/105] + [1/42 + 7/240 + 79/1680] + [1/42 + 9/280 + 37/840] + [1/42 + 59/1680 + 23/560] + [41/1680 + 61/1680 + 11/280] + [23/840 + 17/560 + 71/1680] <---> [2 * 1/42 + 1/28] + [1/42 + 11/420 + 1/30] + [1/42 + 1/35 + 13/420] + [1/28 + 1/21] + [4/105 + 19/420] + [17/420 + 3/70] + [1/42 + 41/1680 + 59/1680] + [1/42 + 23/840 + 9/280] + [1/42 + 7/240 + 17/560] + [61/1680 + 79/1680] + [11/280 + 37/840] + [23/560 + 71/1680] <---> 2 * [3 * 1/42] + [1/42 + 1/21] + [11/420 + 19/420] + [1/35 + 3/70] + [13/420 + 17/420] + [1/30 + 4/105] + [2 * 1/28] + [41/1680 + 79/1680] + [23/840 + 37/840] + [7/240 + 71/1680] + [17/560 + 23/560] + [9/280 + 11/280] + [59/1680 + 61/1680] i got this by joining two of the partitions i found for (5, 6, 7) and dividing by 2 so that all parts add to 1 . it is guaranteed to be a refinement of 10 * 1/10 , 12 * 1/12 and of 14 * 1/14 . then i rearranged the parts to show that it's also a refinement of 8 * 1/8 . note that the largest part is 1/21 , which is twice the smallest part. thus we get another partition by splitting the largest part in half. there are also other ways to rearrange the same set of parts within the 8 * 1/8 . also, it is easy to show that T(5k, 7k) = 1/(21k) , so that 1/42 is not only a lower bound for T(8, 10, 12, 14) ; it's the exact value. mike