Timeline for A generalization of Feit–Thompson conjecture, for square-free integers
Current License: CC BY-SA 4.0
18 events
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| Mar 13, 2022 at 12:32 | comment | added | user142929 | @SylvainJULIEN please I've edited a (my last on MathOverflow) comment in the post related with an attempt to get an interpretation of the sum of divisors function, prime numbers and chirality. It is the post with identificator 407529 , can you think about it? Many thanks. | |
| Mar 13, 2022 at 12:31 | comment | added | user142929 | @Wojowu please I've edited a (my last on MathOverflow) comment in the post with identificator 407529 , can you think about it? Many thanks. | |
| Nov 12, 2020 at 16:30 | comment | added | user142929 | Thank you very much, I appreciate this community of users very much @CarloBeenakker , I hope your colleagues (professors) and users of this website are in good health. | |
| Nov 11, 2020 at 9:57 | comment | added | Carlo Beenakker | just reaching out to you when I read you are losing your internet access from home; we will miss you as a frequent contributor! | |
| S Nov 3, 2020 at 18:00 | history | bounty ended | CommunityBot | ||
| S Nov 3, 2020 at 18:00 | history | notice removed | CommunityBot | ||
| S Oct 26, 2020 at 16:32 | history | bounty started | user142929 | ||
| S Oct 26, 2020 at 16:32 | history | notice added | user142929 | Authoritative reference needed | |
| Apr 15, 2020 at 18:59 | comment | added | Sylvain JULIEN | I don't quite get your point right now but I'll try to think of it and let you know. | |
| Apr 15, 2020 at 17:49 | comment | added | user142929 | I write this comment for your attention @SylvainJULIEN in this and previous comments if you want to think about the previous thread of comments. Additionally I believe that the following conjecture is true: If an integer $n\geq 2$ satisfies $\sigma(\sigma(n))=1+\frac{n\operatorname{rad}(n)-1}{\operatorname{rad}(n)-1}$, then its sum of divisors $\sigma(n)$ is a prime number (here $\operatorname{rad}(n)$ is the product of distinct primes dividing $n$). With the help of Ordowski's claim it is easy to check that each term of the sequence A023194 satisfies previous equation. Many thanks. | |
| Apr 15, 2020 at 7:57 | comment | added | user142929 | (2/2) OEIS that is precisely the integers $n\geq 2$ such that $\sigma(n)=\text{ a prime}$. Finally see the comment in previous A023194 by Thomas Ordowski (Nov 18 2017). I edit these comments with the intention if you @Wojowu or the user in commments want to think about it (my belief is that maybe it is potentially related to compositions of the sum of divisors function with, in a way that I don't know, the Dedekind psi function or the Euler's totient function, and related to constructible polygons with compass and straightedge, Fermat primes and Mersenne primes). I hope don't disturb. | |
| Apr 15, 2020 at 7:57 | comment | added | user142929 | (1/2) Since few months ago my belief is that there exists a "nice" interpretation of the sum of divisors function $\sigma(n)$ that "moves rectangles to rectangles in a special way" (I can not explain better my belief). I consider rectangles $a\times b$ with $a\geq 1$ and for integers $a<b$ formed by identical unitary squares $1\times 1$. See also the recent post MathOverflow 357376 with title Another kind of primality related to tessellations by polygons, which inspires in my opinion that $\sigma(\text{rectangle})=\text{ a convex}$ with the exception of the sequence A023194 from the | |
| Apr 13, 2020 at 15:45 | comment | added | Wojowu | @SylvainJULIEN There are (at least) two issues with this. First of all, how do you want to take an intersection of automorphism groups of distinct rings? Second, $\mathbb Z_p$ always has a trivial automorphism group, so your equivalence doesn't hold since the intersection is also trivial for $p=q$. | |
| Apr 11, 2020 at 19:13 | comment | added | user142929 | Many thanks for your ideas and attention. My abstract algebra isn't the best @SylvainJULIEN | |
| Apr 11, 2020 at 16:06 | comment | added | Sylvain JULIEN | More precisely, maybe we can try to prove that FTC is equivalent to $p\neq q\Longleftrightarrow Aut(Z_{p})\cap Aut(Z_{q})=\{e\}$. | |
| Apr 11, 2020 at 15:49 | comment | added | Sylvain JULIEN | Maybe this could be reduced to the original Feit-Thompson conjecture considering that, writing the set of prime factors of $p$ as $p_i$ and the set of prime factors of $q$ as $q_j$ one has $(p_i,q_j)=1$. The only idea I have so far is to consider automorphisms of the rings of $p$-adic and $q$-adic integers. | |
| Apr 11, 2020 at 9:35 | comment | added | user142929 | To the moderator team of MO/MSE, as soon there is feedback about if this question is interesting for MathOverflow, feel free to handle the MSE 3591614 question as you consider it is right (I say this since I never had a cross-posted question). | |
| Apr 11, 2020 at 9:13 | history | asked | user142929 | CC BY-SA 4.0 |