Timeline for If $N = qn^2$ is an odd perfect number with $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?
Current License: CC BY-SA 4.0
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Sep 11 at 5:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Aug 12 at 4:07 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
edited body
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S Jul 1 at 4:38 | history | suggested | no upstairs |
Tag revision
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Jun 30 at 18:52 | review | Suggested edits | |||
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Jun 11, 2022 at 13:51 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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Jul 9, 2018 at 10:10 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
added GCD constraint
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Feb 12, 2016 at 11:59 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added parentheses
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Feb 7, 2016 at 21:53 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
inserted a brief remark about an error with elementary logic
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Feb 7, 2016 at 12:19 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added further details
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Feb 7, 2016 at 12:11 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added further details
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Feb 6, 2016 at 8:38 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
corrected upper bound for $\sigma(q)/n$
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Feb 6, 2016 at 7:10 | comment | added | Jose Arnaldo Bebita | @SylvainJULIEN, the same results have already been communicated to myself by Patrick starting December 2015. He e-mailed me a partial proof for $q^k < n$ last January 28, and did advise me that he will be posting the same to the arXiv "in the next couple of days". | |
Feb 6, 2016 at 6:57 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
corrected the incorrect upper bound for $1/n$ from ${10}^{-500}$ to ${10}^{-375}$
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Feb 5, 2016 at 20:46 | answer | added | wythagoras | timeline score: 0 | |
Feb 5, 2016 at 14:33 | comment | added | Sylvain JULIEN | Arnie, have you read this: arxiv.org/abs/1602.01591 ? | |
Feb 4, 2016 at 21:00 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added hyperlink to Ochem and Rao's paper in Math. Comp.
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Feb 4, 2016 at 20:24 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
removed reference to arXiv preprint of OP
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Feb 4, 2016 at 20:16 | history | edited | Jose Arnaldo Bebita |
added "divisors-multiples" tag
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Feb 4, 2016 at 20:14 | comment | added | Jose Arnaldo Bebita | The abundancy index of $n$, denoted by $I(n)$, is defined as the ratio $$I(n) = \frac{\sigma(n)}{n}.$$ | |
Feb 4, 2016 at 20:13 | comment | added | Jose Arnaldo Bebita | The quasi-Euler prime of the lone spoof odd perfect number that we know of (i.e., $D = 198585576189$), is the factor $q_1 = 22021$ that has odd exponent and that makes $D$ "perfect", if we pretend that $q_1 = {{19}^2}\cdot{61}$ is "prime". | |
Feb 4, 2016 at 20:09 | comment | added | Jose Arnaldo Bebita | @Seva, the Euler prime of an odd perfect number is the unique prime that occurs to an odd exponent (conjectured to be $1$ by Descartes, Frenicle, and most recently by Sorli). | |
Feb 4, 2016 at 20:06 | comment | added | Seva | What is an "Euler prime"? What is a "quasi-Euler prime"? What is $I(n)$? | |
Feb 4, 2016 at 18:00 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added the detail: "with Euler prime $q$" in the **Question**
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Feb 4, 2016 at 17:40 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
fixed missing parentheses
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Feb 4, 2016 at 17:32 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
added further details
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Feb 4, 2016 at 17:13 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
improved the upper bound for $\sigma(q)/q = 1 + (1/q) < 1 + (1/n)$ from $6/5$ to $1 + {10}^{-500}$
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Feb 4, 2016 at 14:02 | history | asked | Jose Arnaldo Bebita | CC BY-SA 3.0 |