Timeline for Arranging numbers from $1$ to $n$ such that the sum of every two adjacent numbers is a perfect power
Current License: CC BY-SA 3.0
14 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| May 26, 2019 at 0:56 | comment | added | Gerry Myerson | It appears to be proved that for all $N\ge25$ one can arrange all the numbers from 1 to $N$ in a row such that the sum of every two adjacent numbers is a perfect square. See #22 by R. Gerbicz at mersenneforum.org/showthread.php?p=477787 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Aug 3, 2015 at 14:11 | answer | added | castor | timeline score: 6 | |
| Mar 13, 2015 at 3:52 | comment | added | Gerry Myerson | The cube case with $n=305$ and the cyclical cube case with $n=473$ appeared some years ago at primepuzzles.net/puzzles/puzz_311.htm | |
| Mar 12, 2015 at 23:43 | answer | added | Moritz Firsching | timeline score: 12 | |
| Mar 12, 2015 at 11:47 | history | edited | mathlove | CC BY-SA 3.0 |
changed an expression
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| Mar 12, 2015 at 11:47 | comment | added | martin | @mathlove There is some more info on circular loops of square arrangements at oeis.org/A071984 | |
| Mar 12, 2015 at 10:50 | answer | added | martin | timeline score: 16 | |
| Mar 12, 2015 at 0:21 | comment | added | Mirko | it might be that my interpretation of the quantifiers is different from your intention. I take what you say to mean "$\exists n, \forall N$, Condition" whereas judging from your reply to my comment you might perhaps mean "$\forall N, \exists n$, Condition", that is, apparently you mean that $n$ would depend on $N$ (which I now realize is reasonable), whereas the way I read it $n$ is fixed first, and then all positive integers $N$ work for this same $n$, which prompted my objection. If $n$ is intended to depend on $N$ then perhaps this should be made clearer to avoid possible misinterpretation. | |
| Mar 11, 2015 at 21:47 | comment | added | mathlove | @user48481MirkoSwirko: I don't understand why you are comparing $2^N$ with $2n-1$. Since $n$ is different from $N$, we can take sufficiently large $n$, say $n\ge (2N)^N$, for large $N$ | |
| Mar 11, 2015 at 18:49 | comment | added | Mirko | it seems to me you are asking for too much in Q1 with the "for each $N$" part. Clearly the sum of two adjacent numbers is at most $2n-1$, while $m^N$ is at least $2^N$, so for large $N$ the number $2^N$ (and any $m^N$) would clearly overshoot $2n-1$. | |
| Mar 11, 2015 at 12:17 | comment | added | Gerry Myerson | There is some information on the square arrangements at oeis.org/A090460 | |
| Mar 11, 2015 at 11:55 | history | asked | mathlove | CC BY-SA 3.0 |