17 Dec
2017
17 Dec
'17
10 a.m.
My blunder, I just didn't read carefully enough! All seems in order. On Sun, Dec 17, 2017 at 11:54 AM, Tom Karzes <karzes@sonic.net> wrote: > Hi Allan, > > Yes, I had mentioned the fact that each 1/4 must be the sum of exactly > two components at the very beginning, then didn't mention it again: > > 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). > > I also had the same thought about using the LCM restriction earlier to > further contrain the component sizes. I think it's safe to say that > at least three of the component fractions must have a factor of 3 in > their denominator (when reduced to lowest-terms), since each of the > three 1/3 sums must contain at least one such fraction. Further, > each 1/5 sum or 1/4 sum that contains one such fraction must contain > at least one more such fraction, in order to eliminate the factor > of 3. So if there are exactly three such components, then all three > must occur together in a single 1/5 sum, and in a single 1/4 sum. > Otherwise there must be more than three such fractions. > > Tom > > > Allan Wechsler writes: > > 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 > > > > > > > > > > > > > > > > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >