Timeline for Around a characterization for even perfect numbers, similar than Euclides-Euler theorem, in terms of totatives
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 21, 2019 at 8:57 | history | bounty ended | user142929 | ||
| S Nov 21, 2019 at 8:57 | history | notice removed | user142929 | ||
| S Nov 14, 2019 at 18:52 | history | bounty started | user142929 | ||
| S Nov 14, 2019 at 18:52 | history | notice added | user142929 | Draw attention | |
| Nov 14, 2019 at 11:49 | comment | added | user142929 | In my first comment was added that ..since $M$ is perfect..., instead of the right claim ..since $\frac{M+1}{2} M$ is perfect... | |
| Oct 30, 2019 at 15:42 | answer | added | FusRoDah | timeline score: 2 | |
| Oct 30, 2019 at 6:16 | comment | added | user142929 | As aside comment I think that a similar claim and conjecture (a third and last characterization similar than Euclides-Euler theorem) are feasible in terms of the Dedekind psi function $\psi(n)$, in particular the conjecture: An integer $r\geq 1$ satisfies $$\psi\left(\frac{(3r-1)(3r-2)}{2}\right)=\frac{3}{4}(3r-1)^2$$ if and only if $\frac{3r-1}{2}(3r-2)$ is an even perfect number greater than $6$ (being $3r-2=2^p-1$ its associated Mersenne prime). As reference the Wikipedia Perfect number that refers the relationship between even perfect numbers and centered nonagonal numbers. | |
| Oct 28, 2019 at 17:53 | comment | added | user142929 | Thus if you can to prove the conjecture in my Question, then (in particular) the following characterization for even perfect numbers will be feasible: An integer $m\geq 1$ satisfies the identity $$\varphi((m+1)(2m+1))=m(m+1)$$ if and only if $(m+1)(2m+1)$ is an even perfect number (being $2m+1=2^p-1$ its associated Mersenne prime). | |
| Oct 28, 2019 at 17:44 | comment | added | user142929 | All, the comparison with the characterization for even perfect numbers due to Euclides and Euler comes from the case $\lambda=\mu=1$, and from the fact that as a consequence of Euclides-Euler theorem for even perfect numbers $n=2^{p-1}\cdot(2^p-1)$ is a triangular number $n=\frac{M+1}{2}\cdot M$ where $M=2^p-1$ is its associated Mersenne prime. That's $$\sigma\left(\frac{M(M+1)}{2}\right)=M(M+1)$$ since $M$ is perfect. As it was said feel free to provide me the opinion if my Claim and related conjecture are potentially interesting. I am waiting for your feedback, many thanks. | |
| Oct 28, 2019 at 16:05 | history | edited | user142929 | CC BY-SA 4.0 |
Fixed a typo
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| Oct 28, 2019 at 15:59 | history | asked | user142929 | CC BY-SA 4.0 |