Timeline for Is $\liminf \frac{\sigma_{k}(({2}^{m-1})({2^m-1}))}{\phi_{k}(({2}^{m-1})({2^m-1}))}$ finite for every $k$?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| S Jan 8, 2016 at 14:53 | history | bounty ended | CommunityBot | ||
| S Jan 8, 2016 at 14:53 | history | notice removed | CommunityBot | ||
| Jan 5, 2016 at 14:40 | comment | added | Bulois Michael | Wow! Is there a link between the date of the question and the value of $2^{m-1}(2^m-1)$ for $m=6$? | |
| S Dec 31, 2015 at 13:19 | history | bounty started | zeraoulia rafik | ||
| S Dec 31, 2015 at 13:19 | history | notice added | zeraoulia rafik | Draw attention | |
| Dec 31, 2015 at 10:04 | answer | added | Jan-Christoph Schlage-Puchta | timeline score: 3 | |
| Dec 29, 2015 at 23:20 | comment | added | zeraoulia rafik | @GH from MO ,i think also and so it is hard to show the limit above is finit using note (02) | |
| Dec 29, 2015 at 23:16 | comment | added | GH from MO | @zeraouliarafik: I don't think ${\phi_{k}(({2}^{m-1})({2^m-1}))}=({2}^{m-1-k})\phi_{k}({2^m-1})$ is true in general. The iterated $\sigma$ and $\phi$ functions are not multiplicative. You can probably find a counterexample by plugging in small values of $m$ and $k$. | |
| Dec 29, 2015 at 22:04 | comment | added | zeraoulia rafik | in note(02) , just i would like to write my limit in other form for using shinzel and sieve conjecture using simplification which montioned in note 2 " using property of multiplicative function and i asked also in this note if it is true for all m, and k ? | |
| Dec 29, 2015 at 21:20 | answer | added | Gerhard Paseman | timeline score: 0 | |
| Dec 29, 2015 at 18:37 | comment | added | GH from MO | Personally, I think this question is beyond current technology. | |
| Dec 29, 2015 at 16:36 | comment | added | GH from MO | The "proof theory" tag should be reserved to (certain) questions in mathematical logic, namely those which concern formal proofs and their theory. | |
| Dec 29, 2015 at 16:35 | history | edited | GH from MO |
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| Dec 29, 2015 at 11:40 | comment | added | zeraoulia rafik | and it is the same for what i wrote but what about more iteration of k ? | |
| Dec 29, 2015 at 11:39 | comment | added | Meysam Ghahramani | For all $m\geq 1$ we have $gcd(2^{m-1},2^m-1)=1$. | |
| Dec 29, 2015 at 11:20 | history | edited | zeraoulia rafik | CC BY-SA 3.0 |
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| Dec 29, 2015 at 11:03 | history | edited | zeraoulia rafik | CC BY-SA 3.0 |
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| Dec 29, 2015 at 10:58 | history | asked | zeraoulia rafik | CC BY-SA 3.0 |