Timeline for Is there a fixed integer $x>1$ satisfing ${\sigma}^{k}(x)\equiv 0\pmod{x}$ for all positive integers $k$?
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29 events
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| Jan 1, 2016 at 0:00 | comment | added | zeraoulia rafik | @GerhardPaseman, The solution of that problem in this paper :Dieter Bode, Über eine Verallgemeinerung der vollkommenen Zahlen, Dissertation, Braunschweig, 1971. , but i can't get it , and it's stated that no even m-superperfect existed for m\geq 3 | |
| Aug 2, 2015 at 15:01 | comment | added | zeraoulia rafik | @GerhardPaseman,This type of problems is beyond the actual theory.and should study the superperfect number not metaperfect number . | |
| Jul 21, 2015 at 0:51 | comment | added | zeraoulia rafik | @GerhardPaseman, ok thank you very much for this observation | |
| Jul 21, 2015 at 0:47 | comment | added | Gerhard Paseman | @zeraouliarafik , I recommend that you no longer comment on this answer; it is getting too long to follow. Also, you should check that such n do not satisfy your condition: \sigma(114^(114k)) is odd. Further, I do not want to be emailed on conjectural statements. You should show that you put in some work to resolve a question you have. Gerhard "Look At Twice A Square" Paseman, 2015.07.20 | |
| Jul 21, 2015 at 0:31 | comment | added | zeraoulia rafik | @GerryMyerson, you can check also this number :$n={114}^{114k}$, for $k$ is a positive integer . | |
| Jul 20, 2015 at 16:43 | comment | added | zeraoulia rafik | @joro, I understand according to ur above comment that it's impossible to get a such number X satasfing the problem | |
| Jul 20, 2015 at 16:38 | comment | added | joro | @zeraouliarafik I don't think so. The inequality is unconditional and RH is equivalent to another inequality. | |
| Jul 20, 2015 at 16:17 | comment | added | zeraoulia rafik | well , then the question abouve may equivalent to Riemann Hypothesis according to your comments | |
| Jul 20, 2015 at 16:13 | comment | added | joro | It is known that sigma(n)/n can't grow too fast: en.wikipedia.org/wiki/Divisor_function#Approximate_growth_rate Robin also proved, unconditionally, that the inequality $\ \sigma(n) < e^\gamma n \log \log n + \frac{0.6483\ n}{\log \log n} $ holds for all n ≥ 3. | |
| Jul 17, 2015 at 21:56 | comment | added | zeraoulia rafik | @GerhardPaseman, I have send to you some related papers to this problem "check email" !!!! | |
| Jul 17, 2015 at 4:02 | comment | added | Gerhard Paseman | I recommend further discussion in email, not comments. Also, it is unclear how the paper relates to the problem. Iterates of the totient function are not necessarily related to iterates of the sigma function. Gerhard "You Can Explain In Email" Paseman, 2015.07.16 | |
| Jul 17, 2015 at 1:24 | comment | added | zeraoulia rafik | @GerhardPaseman: pleas look this:Theorem 2.3 :cs.uwaterloo.ca/journals/JIS/VOL9/Shparlinski/shpar43.pdf, i think it works with integer 2 | |
| Jul 16, 2015 at 21:55 | comment | added | zeraoulia rafik | @GerhardPaseman, I read many papers about multiperfect number but i think this is make a new problem ,i can't find any equivalence to this question " it is very hard" | |
| Jul 16, 2015 at 21:53 | comment | added | Gerhard Paseman | I think you should not accept this version. I am thinking about a proof, as well as related questions (which you should think about as well). If there is no further progress, I will add a further edit which says what difficulties there are in showing that such a sequence exists. If you find the result acceptable, it would be OK with me to accept that future version. Gerhard "Now Back To The Present" Paseman, 2015.07.16 | |
| Jul 16, 2015 at 21:49 | comment | added | zeraoulia rafik | @GerhardPaseman, can I accept the answer or should be waite the complet proof ? | |
| Jul 16, 2015 at 0:40 | comment | added | zeraoulia rafik | @GerhardPaseman , sorry i don't know i deleted it excuse me | |
| Jul 16, 2015 at 0:33 | comment | added | Gerhard Paseman | @zeraouliarafik: please delete your last comment. I present my personal information the way I do to keep a low profile. Gerhard "Doesn't Need Spambots Knowing It" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 23:54 | comment | added | Gerhard Paseman | @zeraouliarafik for permission to email me on this, you have it. For the email address, you have to do (a small amount of) research. Gerhard "Hint: Try The User Page" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 23:25 | comment | added | Gerry Myerson | $\sigma^{32}(2)=564210119465811\equiv1\bmod2$. I don't think it has been proved that $\sigma^k(2)\equiv0\bmod2$ for all $k\ge33$. Data at oeis.org/A007497/b007497.txt | |
| Jul 15, 2015 at 22:44 | comment | added | zeraoulia rafik | @GerhardPaseman, ok no problem , give me your Email to i can contact you and i'm just in the paper as Author who stated the problem . | |
| Jul 15, 2015 at 22:25 | comment | added | Gerhard Paseman | @zeraouliarafik, I'm not sure. I would be happy to be acknowledged for my contributions should you wish to write such a paper. I can be pretty picky about style in something that has my name as author though. However, people have studied iterates of $\sigma()$ before; If you want my help with such a paper, my first question would be to get the bibliography of papers you have read on the subject, and my second would be for the list of papers you have skimmed or intend to skim. Gerhard "Acknowledging The Literature Is Important" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 22:19 | comment | added | zeraoulia rafik | @GerhardPaseman, if you see it that is not standard and make a new problem , can we do a paper about this topic ? | |
| Jul 15, 2015 at 22:17 | comment | added | Gerhard Paseman | @zeraouliarafik But that is a different question: you are now asking for an integer m and integer x such that $x$ divides $\sigma^k(m)$ for all $k$. It may be possible that 2 is such an integer, but even that is not known. Gerhard "This Question Is Hard Enough" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 21:54 | comment | added | zeraoulia rafik | @GerryMyerson, if you seek to the asymptotics integer , i think 114 mod 6 is good example and fails only for k=9,12,18 | |
| Jul 15, 2015 at 21:29 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
added 1870 characters in body
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| Jul 15, 2015 at 20:53 | comment | added | Gerhard Paseman | I am now confident the answer to the question is no; the basic reason is that the function $r(m)=\sigma(m)/m$ "can't grow fast enough" to support the existence of such a number. I will attempt an argument. Gerhard "Sleeping On It Is Good" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 7:27 | comment | added | Gerhard Paseman | That is smaller than I would have expected. Gerhard "How About One More Iterate?" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 3:04 | comment | added | Gerry Myerson | The number $n=13188979363639752997731839211623940096$ satisfies $\sigma(n)=5n$ and, since $\gcd(5,n)=1$, $$\sigma^2(n)=\sigma(5n)=6\sigma(n)=30n$$ so at least there's one example where $m_1$ and $m_2$ are multiples of $m$. Whether $n$ qualifies as "very large indeed" is a matter of taste. | |
| Jul 15, 2015 at 1:16 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |