Tom, at first I thought I saw a gap in this logic, but then I realized that in the 4 * 1/4 assembly, all of the quarters must be divided into exactly two pieces. Having realized that lemma, I reread your reasoning and everything made sense. It feels like the LCM machinery you bring in at the very end might be leveraged to carry more of the proof. Now we just have to show f(5) != 9. On Sun, Dec 17, 2017 at 10:41 AM, Tom Karzes <karzes@sonic.net> wrote:
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
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