Timeline for Conjecture on signed sum of integer fractions x/y from 1..N?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2011 at 23:28 | comment | added | Gerhard Paseman | Two points: a) there are N(N-1) ways of writing u_i, but there are (for N>3) less than N(N-1) values for the u_i. I apologize for the conflation. b) There are N! 2^(ceil(N/2)) ways of writing the sum, which because of associativity and commutatitvity will repeat each sum (ceil(N/2))! times. Unless you choose some subset of these ways to investigate, it is not clear that you will avoid subsums of the form 1/a - b/ab for 1<a<b<sqrt(N), and thus there will be further repats of sums. Also, I will stop commenting here. Gerhard "NathOverflow is Bad For Discussions" Paseman, 2011.01.28 | |
| Jan 28, 2011 at 9:46 | comment | added | smci | >I also think that ordering will not solve the problem of duplicate sums. I disagree. It removes most of them. I actually programmed the recursion. Give me a counterexample if you still think so? Here's a probabilistic rationale: The larger N gets, the smaller the probability that any two u_i (when numerator and denominator are randomly assigned) will be equal. Hence probabilistically, most of the signed sums Σ σ_i (x_i/y_i) should be distinct. This guarantees us a denser set of results to choose from. | |
| Jan 28, 2011 at 9:41 | comment | added | smci | >My concern arises from your statement that there are N!(N-1)! values for u_i. I think there are N(N-1) values for the (as yet signed) u_i. You're right, that statement was sloppy. Consider N=2M (even N), and we choose ordering by increasing denominator. Then there are (N/2)! ways of choosing the denominators {y_i}, and also (N/2)! ways of choosing the numerators {x_i}. I think this eliminates double-counting. In total (N/2)! = (M!)^2 ways of choosing the {u_i}. | |
| Jan 28, 2011 at 1:49 | comment | added | Gerhard Paseman | My concern arises from your statement that there are N!(N-1)! values for u_i. I think there are N(N-1) values for the (as yet signed) u_i. I also think that ordering will not solve the problem of duplicate sums. Indeed one might use the fact along with the density of primes to argue that there will be finitely many N for which S(N)= 0. You still might consider an incremental definition of T(N). Gerhard "Ask Me About System Design" Paseman, 2011.01.27 | |
| Jan 27, 2011 at 21:38 | comment | added | smci | In fact maybe another "most legible" ordering principle is "in order of the term containing the next lowest number, whether in numerator or denominator". So e.g. S(7) = 0 = 1/3 +5/2 -4 +7/6 However this is way harder to program recursively than "denominators {y_i} in increasing order". S(7) = 0 = -4 +5/2 +1/3 +7/6 Yet a fourth "well-ordering principle" would be "in decreasing order of absolute magnitude of each term u_i = (x_i/y_i)" S(7) = 0 = -4 +5/2 +7/6 +1/3 This is visually nicest, but implies an exponential number of sorts Can we first agree on what ordering is best for our purposes? | |
| Jan 27, 2011 at 21:28 | comment | added | smci | In fact a better ordering principle is probably denominators {y_i} must be in increasing order. This makes for better legibility of the solution, but programming the recursion gets a little bit more annoying. | |
| Jan 27, 2011 at 21:23 | comment | added | smci | Dude, I wrote that u_i = (x_i/y_i) is one term, σ_i is its sign, and the summation is Σ σ_i (x_i/y_i) Yes, rarely some of the sums are equal, if we adopt some (arbitrary) ordering principle such as e.g. require the numerators {x_i} to be in increasing order. This throws out duplicate sequences with terms swapped. After that there is nearly-zero probability of two different sequences having the same sum. So the probabilistic argument gives us a (quadratic?) range of choices for each term. | |
| Jan 19, 2011 at 22:39 | comment | added | Gerhard Paseman | Since you asked, my reaction is that it is illformed. Is u_i one of the fractions or the sum? For a given 2m, there are less than 4m^2 such fractions allowed, and you don't want the sum of any of them, but of a particular subset of them which is challenging to enumerate. Until you clean up that part, I am not encouraged to proceed to the estimate of the enumeration, much less to the argument. Also, it is likely that some of the sums will be equal, and it is not clear how the probabilistic argument will handle that. Gerhard "Ask Me About System Design" Paseman, 2011.01.19 | |
| Jan 19, 2011 at 22:18 | comment | added | smci | Even if you can't prove the inductive step, any reaction to my suggestion on a probabilistic proof? | |
| Jan 19, 2011 at 17:51 | comment | added | Gerhard Paseman | I'm surprised, as I hadn't noticed the negative vote. Also, (I think it was) your observation about s(primorial(n))>= n was useful in the happy prime new year thread, and I was sorry to see it go. Perhaps it will return. Gerhard "Thank You For Your SUpport" Paseman, 2011.01.19 | |
| Jan 18, 2011 at 23:30 | history | answered | Gerhard Paseman | CC BY-SA 2.5 |