This is so similar in feel to the "muffin problem" that we talked about a couple of years ago, that I am convinced there must be a relationship between the problems. But I can't recall the original formulation of the muffin problem. I think we can use reasoning like Tom Karzes's to show that f(5) > 7. Five fifths must be possible, and if we have only 7 pieces, then when the five fifths are arranged, there can be only two internal divisions. If those are in the same fifth, then we can't achieve four quarters; if they are in different fifths, then the three atomic fifths must be combined with 1/20 to get to 1/4, and there's no way to generate the required three 1/20 pieces. On Fri, Dec 15, 2017 at 12:50 PM, Tom Karzes <karzes@sonic.net> wrote:
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
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