I believe I can prove that f(5) is not 8, which leaves either 9 or 10. Proof: If we assume f(5) is 8, then in the 4 * 1/4 case, each 1/4 must be the sum of exactly two smaller component fractions (since the largest possible component fraction is 1/5). Now enumerate the possible component breakdown in the 5 * 1/5 case. The possibilities are: (a) 2, 2, 2, 1, 1 (b) 3, 2, 1, 1, 1 (c) 4, 1, 1, 1, 1 These are all of the ways to divide 5 values into 8 components. In each case, consider what values are required in order to combine these 8 values into 4 * 1/4. In case (a), exactly two of the component values are 1/5. That means two others must be 1/20 in order to make 1/4. 1/5 - 1/20 is 3/20. So we have two instances of (3/20 + 1/20) and two instances of (1/5). That leaves a single pair of values. After pairing (1/5 + 1/20) twice to get 2 * 1/4, we are left needing 2 * 1/4 out of 2 * 3/20 and two other values, which must therefore be 2 * 1/10 since (3/20 + 1/10) is 1/4. So for case (a) we have: 2 * 1/5, 2 * 3/20, 2 * 1/10, 2 * 1/20 In case (b), exactly three of the component values are 1/5. That means that three others must be 1/20 in order to make 1/4. So three other values must 1/20 in other to make 1/4. The only way to get 3 * 1/20 out of the remaining five components, three of which sum to 1/5 and two others of which sum to 1/5, is for one to be (1/10 + 1/20 + 1/20) and the other to be (3/20 + 1/20). So for case (b) we have: 3 * 1/5, 1 * 3/20, 1 * 1/10, 3 * 1/20 In case (c), exactly four of the component values are 1/5. That means that four others must be 1/20 in order to make 1/4. So for case (c) we have: 4 * 1/5, 4 * 1/20 That covers all possible ways to make both 5 * 1/5 and 4 * 1/4 out of 8 component values. However, so far this has completely ignored the need to make 3 * 1/3 from the component values. But this is clearly impossible, since in all cases the LCM of the denominators of the components is 20, and x/20 cannot be 1/3 for integer x. So this rules out f(5) being 8, leaving only 9 and 10 as candidates. It may be possible to rule out 9 using a similar but more convoluted argument, but I have not attempted to do so. Tom Tom Karzes writes:
That's right. It's definitely one of 8, 9, or 10. I suspect it's 10, but I can't prove it isn't 8 or 9.
Tom
Allan Wechsler writes:
If I've been following the bidding up until now, we still don't know whether f(5) is 8, 9, or 10.
On Sat, Dec 16, 2017 at 5:59 PM, Tom Karzes <karzes@sonic.net> wrote:
I believe an upper bound for f(n) is given by https://oeis.org/A092249 which can be interpreted as the number of segments resulting from slicing the interval [0, 1] into 2, 3, ... n uniform segments. Clearly this satisfies all of the required sums. The first several values are:
1,2,4,6,10,12,18,22,28,32,42,46,58,64,72,80,96,...
Note that this is also https://oeis.org/A002088 but without the initial zero entry.
The question is, is it ever possible to do better than this?
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