I believe f(5) is at most 10. Here's a 10-value solution for n=5: 3 * 1/5 2 * 1/12 4 * 1/20 1 * 1/30 These can be combined as: 5: 1/5 1/5 1/5 4 * 1/20 2 * 1/12 + 1/30 4: 1/5 + 1/20 1/5 + 1/20 1/5 + 1/20 2 * 1/12 + 1/20 + 1/30 3: 1/5 + 1/12 + 1/20 1/5 + 1/12 + 1/20 1/5 + 2 * 1/20 + 1/30 Since 2 is a factor of 4, it's obtainable by grouping pairs from the 4-slice solution. Can anyone improve this? Tom Tom Karzes writes:
I think f(4) has to be 6. Consider that 1/4 + 1/4 + 1/4 + 1/4 must be possible. If f(4) is 5, then that means three of those 1/4 are indivisible, and the fourth is divisible once, into two fractions that sum to 1/4. So even if those two fractions are added to two of the other 1/4 values, it still leaves one bare 1/4 (and the other two couldn't both be increased to 1/3 anyway).
With 6 values, it's easy: 1/12, 1/12, 1/12, 1/4, 1/4, 1/4 works.
Tom
James Propp writes:
Perhaps my question has been considered in the past as a question about cutting an interval into pieces, since the circularity of the pie/pizza/whatever is irrelevant. (The very first radial cut effectively turns a problem about cutting a disk into wedges into a problem about cutting an interval into subintervals.)
Or maybe we should get rid of geometry entirely, and just ask: What is the smallest collection of fractions (with repetitions allowed), summing to 1, such that by combining fractions in the collection we can write 1 as 1/2 + 1/2, or as 1/3 + 1/3 + 1/3, or ..., or as 1/n + 1/n + ... + 1/n?
Let f(n) be the smallest possible number of such fractions. Clearly f(1) = 1 and f(2) = 2, and it's not hard to show that f(3) = 4. I haven't figured out f(4) (it's either 5 or 6). Has anyone seen this sequence before?
Jim