Muffin (4,5,6) >= 7/120: 120/6 = 2[7.13], 2[8.12], 2[9.11] 120/5 = 2[7.8.9], 2[11.13], [12.12] 120/4 = 2[7.11.12], 2[8.9.13] Note muffin(5,6) = 1/15, muffin(4,6) = 1/12, muffin(4,5) = 3/40. 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] I believe this partition is unique up to piece size and multiplicity, and that muffin(4,5,6,7) < 1/21. Could muffin(4k,5k,6k,7k) = 1/21 for some k? - Scott

Subject: [math-fun] muffin(2,3,5)

Taking up an earlier suggestion about looking for corefinements of 1 into several sets of unit fractions ... Define Muffin(a,b,...) be the minimum piece size of a corefinement of 1 into pieces of size 1/a, and of 1/b, etc. Muffin(2,3,5) seems to be the smallest interesting case. I get M(2,3) = 1/6, M(2,5) = 1/10, and M(3,5) = 1/12. The M(3,5) split, normalized to 60ths, is 4*5 + 2*6 + 4*7. Thirds: 20 = 4*5 = [2]*(6+7+7). Fifths: 12 = [4]*(5+7) = 6+6. This also works for M(2,3,5): 30 = [2]*(5+5+6+7+7). So M(2,3,5) = 1/12.

Rich