Timeline for Some divisibility constraints in Frobenius coin problem
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2015 at 12:13 | comment | added | Kim Morrison | I think this comment thread has gone on long enough. | |
| Dec 3, 2015 at 2:46 | vote | accept | Turbo | ||
| Dec 3, 2015 at 2:46 | |||||
| Dec 2, 2015 at 20:12 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
Hopefully this suffices.
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| Dec 2, 2015 at 20:06 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
Hopefully this suffices.
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| Dec 2, 2015 at 16:34 | comment | added | Gerhard Paseman | How many integers are there between 2b+1 and 2*(2b+1) - 2 - (2b+1)? Does this help you determine if (2,2b+1) is a reasonable coprime pair? Gerhard "Are You Analyzing At All?" Paseman, 2015.12.02 | |
| Dec 2, 2015 at 7:52 | comment | added | Turbo | I do not understand your insight on $(2,2b+1)$ | |
| Dec 2, 2015 at 7:52 | comment | added | Turbo | I think there are plenty of these pairs. I would not be surprised if for every $n>n_0$ and every $a\in[n,2n]$, there is a $b\in[n,2n]$ or alteast $\log n$ such pairs | |
| Dec 2, 2015 at 5:04 | comment | added | Gerhard Paseman | The correct option is the one you have verified yourself. I assert (2,2b+1) is a reasonable coprime pair. Can you verify or refute that assertion? I say similar things about pairs (3,b) and (4,b). Can you verify those? Can you make headway on (5,b) and (6,b)? Why should (15,b) never be a reasonable coprime pair? Notice I have never proved that there are finitely many reasonable coprime pairs, only predicted that (and later retracted). I have posted things to help you solve your problem, not solve it for you. Gerhard "Believes In Individual Independent Verification" Paseman, 2015.12.01. | |
| Dec 2, 2015 at 3:52 | comment | added | Turbo | In your post you say you have finitely many possibilities only.. in comments you say infinite... which is correct? | |
| Dec 2, 2015 at 2:43 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
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| Dec 2, 2015 at 1:21 | comment | added | Turbo | You are giving in bits and pieces. It is hard to tell what state. | |
| Dec 2, 2015 at 1:09 | comment | added | Turbo | I do not understand ' Then $2s, 3s, 6s$ are also representable, so by your condition $7(a+b)>ab$' | |
| Dec 2, 2015 at 1:08 | history | edited | Turbo | CC BY-SA 3.0 |
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| Dec 2, 2015 at 0:19 | comment | added | Gerhard Paseman | You seem to be ignoring or not comprehending the point I make above. So, let me ask you: Are 2b, 3b, and 6b representable? If they are, for what values of a can you exclude them (otherwise I do not see a and b forming a reasonable pair)? If 2b, 3b, and 6b aren't representable, why not? Gerhard "You Are Meant To Think" Paseman, 2015.12.01 | |
| Dec 2, 2015 at 0:11 | comment | added | Turbo | So are you saying $n<14$ is the highest $n$? Are you implying there is no such $n_0$? | |
| Dec 2, 2015 at 0:08 | comment | added | Gerhard Paseman | $a \lt 7$ unless $b=8$. Gerhard "And Similarly For Six, Five" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 23:33 | comment | added | Gerhard Paseman | Of course not. It is a prediction that turned out to be wrong. If my post and comments don't help you answer your question, I don't think I will be able to enlighten you. Gerhard "Consider The Source Of Noncomprehension" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 22:46 | comment | added | Turbo | ' I predict there will be only finitely many such pairs' is not a proof | |
| Dec 1, 2015 at 22:45 | comment | added | Gerhard Paseman | And in case you missed it, $a \lt 7$. So $n$ can't be arbitrarily large. Gerhard "I Mean The Second N" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 22:43 | comment | added | Gerhard Paseman | $b$ in $(3,b)$ is a number greater than and coprime to $3$, in which case one is looking at representability of numbers in $(b, 2b-3)$ which gives no numbers in that interval with proper divisors in that interval. Looking at slightly larger intervals for larger values of $a$ starts to show the possibilities. So there are infinitely many pairs, but many of them aren't interesting from a divisibility lattice standpoint. Gerhard "Been Staring At Lattices Lately" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 22:40 | comment | added | Turbo | I think there are sufficiently many in a partcular $[n,2n]$ if one exists in that particular $[n,2n]$ | |
| Dec 1, 2015 at 22:32 | comment | added | Gerhard Paseman | Oops. (2,2k+1) works, although somewhat trivially, as does (3,b) for somewhat similar reasons. I think there is little hope for (6,b) though. Gerhard "Done With This For Now" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 22:24 | comment | added | Gerhard Paseman | Indeed, choices are slim if you have $b \lt 2ka \lt 3ka \lt 6ka \lt ab -a-b$ for some positive integer $k$. Gerhard "Finite Is Looking Really Good" Paseman, 2015.12.01. | |
| Dec 1, 2015 at 22:16 | comment | added | Gerhard Paseman | Indeed, for $a\lt b$, what are your plans for $6b$? Gerhard "Maybe This Is Even Simpler" Paseman, 2015.12.01 | |
| Dec 1, 2015 at 22:12 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |