Timeline for Problem related to Frobenius coin problem
Current License: CC BY-SA 3.0
31 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Dec 3, 2015 at 8:55 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 3, 2015 at 4:13 | comment | added | Turbo | @Seva Also to note mathoverflow.net/questions/225047/… is stronger version | |
| Dec 2, 2015 at 8:20 | history | edited | Seva | CC BY-SA 3.0 |
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| Dec 2, 2015 at 8:20 | comment | added | Turbo | @Seva Just one thing (may be I missed before) I think 'some non-negative integer x and y' has to be 'some positive integers x and y' | |
| Dec 2, 2015 at 8:20 | comment | added | Seva | I took the liberty to edit your question; hope it matches what you actually had in mind. | |
| Dec 2, 2015 at 8:18 | history | edited | Seva | CC BY-SA 3.0 |
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| Dec 2, 2015 at 8:06 | comment | added | Turbo | @Seva At most one pair should be represented | |
| Dec 2, 2015 at 7:52 | comment | added | Seva | If both $rs$ and $tu$ are represented, then at least one of $rt,su,ru,st$ should be avoided? Or at least one of $rt,su$, and also at least one of $ru,st$? That is, you allow at most one pair or at most two pairs to be represented? | |
| Dec 1, 2015 at 21:04 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 21:02 | comment | added | Turbo | @seva it is slightly more technical. There are three pairs $(rs,tu)$, $(rt,su)$ and $(ru,st)$. If it represents one pair then it should avoid at least one member of other pairs. This is what I imply. | |
| Dec 1, 2015 at 21:00 | comment | added | Seva | It still seems to me that your condition can be stated in a simpler way: namely, whenever $r,s,u$ and $v$ are positive integers such that the products $rs, uv, ru$, and $sv$ are all smaller than $g(a,b)$, they cannot be all represented by the form $ax+by$. (The condition with $rs,uv,rv$ and $su$ is identical to the one above and can be omitted.) Is this correct? Also, since you do not have any examples of good pairs, it makes sense to ask whether they exist at all in the first place. | |
| Dec 1, 2015 at 20:18 | comment | added | Turbo | @Seva Simplified the writeup further. | |
| Dec 1, 2015 at 20:17 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 8:28 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 8:18 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 8:17 | comment | added | Turbo | @StevenStadnicki Yes, for all values for which lhs holds rhs should hold here | |
| Dec 1, 2015 at 8:16 | comment | added | Steven Stadnicki | @Turbo You should really clarify the quantifiers in your original question, then - as it currently stands, the natural reading is 'if the condition on the left side of the implication arrow holds, then the condition on the right side holds' (this is what an implication usually means!) which is generally equivalent to saying 'for ALL values which satisfy the condition on the LHS, the condition on the right holds'. | |
| Dec 1, 2015 at 8:12 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 8:10 | comment | added | Turbo | @StevenStadnicki $r,s,t,u$ can be such that each of $rs,tu,rt,su,ru,st<g(a,b)$ holds. But I feel pretty bad I did not explicitly state this. Clarified this. Ofcourse what I state will not hold if numbers exceed $g(a,b)$. | |
| Dec 1, 2015 at 7:56 | comment | added | Steven Stadnicki | Doesn't the Frobenius problem itself immediately imply that the answer to your question is negative? All numbers $n$ greater than $(a-1)(b-1)$ have a representation $n=av+bw$ with $v,w$ positive, so as soon as your $r,s,t,u$ are sufficiently large (or more simply as soon as their products are) you'll inevitably have a positive representation. | |
| Dec 1, 2015 at 7:41 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 1, 2015 at 7:35 | comment | added | Seva | Well, I am afraid not quite: having spend some time, I am not sure I got right your definition of a good pair. The four conditions that you mention (if $av+bw=rt$ then $uw<0$ etc) all seem completely equivalent to me as you can switch $r$ with $s$, and $t$ with $u$; am I right? Also, do you have any examples of good pairs? | |
| Dec 1, 2015 at 6:33 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| Nov 30, 2015 at 23:08 | comment | added | Turbo | @Seva you think it is comprehensible now? I do not have an example for good pairs. | |
| Nov 30, 2015 at 21:30 | history | edited | Turbo | CC BY-SA 3.0 |
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| Nov 30, 2015 at 21:24 | history | edited | Turbo | CC BY-SA 3.0 |
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| Nov 30, 2015 at 21:16 | history | edited | Turbo | CC BY-SA 3.0 |
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| Nov 30, 2015 at 21:04 | comment | added | Turbo | @Seva Actually I thought about it, I am not sure $n_0$ has a good upper bound at all. | |
| Nov 30, 2015 at 20:01 | comment | added | Seva | A piece of advice. To receive proper attention and get answered, a problem must be presented in a comprehensible way. For your specific question, you can consider, for instance, defining the notion of a "good pair" $(a,b)$, presenting and explaining some examples of good and "bad" pairs, and only then asking whether there is a good pair in any interval $[n,2n]$. | |
| Nov 30, 2015 at 19:30 | history | asked | Turbo | CC BY-SA 3.0 |