Timeline for Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers
Current License: CC BY-SA 4.0
37 events
| when toggle format | what | by | license | comment | |
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| Oct 12, 2020 at 19:36 | review | Close votes | |||
| Oct 17, 2020 at 3:06 | |||||
| S Oct 4, 2020 at 11:00 | history | bounty ended | CommunityBot | ||
| S Oct 4, 2020 at 11:00 | history | notice removed | CommunityBot | ||
| Sep 30, 2020 at 4:43 | comment | added | mathoverflowUser | @MaxAlekseyev: You are right, it is a way of testing perfectness for an odd number. | |
| Sep 29, 2020 at 16:14 | comment | added | Max Alekseyev | @stackExchangeUser: First, $f_n(x)$ being irreducible for some $n$ does not contradict the existence of an OPN. Second, it's not an attack on OPN, but just an obscure way to test the perfectness of $n$; and it does not have any benefits over the direct testing of the equality $\sigma(n)=2n$. | |
| Sep 29, 2020 at 7:10 | comment | added | mathoverflowUser | @MaxAlekseyev: Using your technique with the linear dependence it seems it is not possible to derive formulas for $a_i$. Hence to evaluate $f_n(x)$ it seems there is not a simple formula since it has coefficients $a_i$ not $A_i$. But I have tested that $f_n(x)$ is irreducible for a lot of odd $n$s, which contradicts $f_n(n)=0$ for an odd perfect number. Thus it still is an attack on OPN. | |
| Sep 28, 2020 at 14:43 | comment | added | Max Alekseyev | @stackExchangeUser: I do not see how these polynomials can help for the OPN problem. To evaluate the polynomials for a given $n$, one should be able to compute $\sigma_3(n)$, and thus be able to compute $\sigma(n)$ and test the quality $\sigma(n)=2n$ directly. | |
| Sep 28, 2020 at 9:38 | comment | added | mathoverflowUser | Dear @MaxAlekseyev: My question is, if your replace $a_i$ and $A_i$ in the question with your identities, then what are the polynomials $F_n(x)$ and $G_n(x)$ which the odd perfect number is a root of? ( Then one can evaluate those polynomials at odd numbers $m$ and see if they have $\gcd=1$. If this is the case, one would have an attack on the OPN problem). Thank you for your help and patience. | |
| Sep 28, 2020 at 9:30 | comment | added | Max Alekseyev | $\gcd(F_n(x),G_n(x))=1$ together with $F_n(n)=G_n(n)=0$ is not possible, since $F_n(n)=G_n(n)=0$ implies that $x-n$ divides $\gcd(F_n(x),G_n(x))$. | |
| Sep 28, 2020 at 9:27 | comment | added | mathoverflowUser | @MaxAlekseyev I think that one can use your identities from the previous question, to replace $a_i$ and $A_i$ with $\sigma_3(n)$. This would lead to 2 polynomials $F_n(x)$ and $G_n(x)$ which hopefully have numerically $\gcd(F_n(x),G_n(x))=1$ and $F_n(n)=0, G_n(n)=0$ for the odd perfect number $n$. I think this would be helpful to have. | |
| Sep 28, 2020 at 9:21 | comment | added | Max Alekseyev | It is not clear what you're asking, unfortunately. | |
| Sep 28, 2020 at 8:48 | comment | added | mathoverflowUser | @MaxAlekseyev: You are right with your observation. But it would be nice if one can use your identities to express $f_n(x)$ and $g_n(x)$ in a different form. Maybe this is more tractable? | |
| Sep 28, 2020 at 8:40 | comment | added | Max Alekseyev | I'm not sure what is the question then here. | |
| Sep 28, 2020 at 8:39 | comment | added | mathoverflowUser | @MaxAlekseyev: No, of course not, since this is off-topic ;) But maybe with your identities we can replace the $a_i,A_i$ with some other untractable function such as $\sigma_3(n)$ etc. ? | |
| Sep 28, 2020 at 8:36 | comment | added | Max Alekseyev | Showing that $\gcd(f_n(x),g_n(x))=1$ for all $n$ would disprove the existence of an odd perfect number, since such $n$ would be a zero of both $f_n(x)$ and $g_n(x)$, meaning that $\gcd(f_n(x),g_n(x))\ne 1$. So, are you asking to disprove the existence of an odd perfect number? | |
| Sep 28, 2020 at 6:38 | comment | added | mathoverflowUser | related question: mathoverflow.net/questions/372766/… | |
| Sep 27, 2020 at 20:24 | comment | added | mathoverflowUser | Using the Ramanujan Identity ( en.wikipedia.org/wiki/Eisenstein_series#Ramanujan_identities ) it follows that every perfect number $n$ satisfies: $8n^4-2n^3-3 \sigma_3(n)n^2+24 A_2=0$. | |
| S Sep 27, 2020 at 13:34 | history | suggested | mathoverflowUser | CC BY-SA 4.0 |
added new conjecture
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| Sep 27, 2020 at 13:14 | comment | added | Jeroen Noels | @JoshuaZ All 38 odd abundant numbers listed in oeis.org/A005231 give a negative f(n,n). I do not know if this generalizes though. | |
| Sep 27, 2020 at 13:09 | comment | added | mathoverflowUser | Dear @JeoenNoels, Thank you for your suggestion. I have edited the question, but the change will appear after someone has approved it. Meanwhile, you maybe might want to look at math.stackexchange.com/questions/3839696/… | |
| Sep 27, 2020 at 7:55 | review | Suggested edits | |||
| S Sep 27, 2020 at 13:34 | |||||
| Sep 26, 2020 at 20:33 | comment | added | JoshuaZ | @stackExchangeUser f(n,n)>0. | |
| Sep 26, 2020 at 17:55 | comment | added | mathoverflowUser | @JoshuaZ: You mean a counterexample to $f(n,n)>0$ or do you mean a counterexample to $f(n,t)$ being irreducible? Thanks for your help. | |
| Sep 26, 2020 at 17:43 | comment | added | JoshuaZ | @JeroenNoels Presumably 945 is a counterexample because it is abundant. In general any abundant number should be a counterexample I think? | |
| S Sep 26, 2020 at 17:29 | history | suggested | mathoverflowUser | CC BY-SA 4.0 |
added example
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| Sep 26, 2020 at 16:28 | review | Suggested edits | |||
| S Sep 26, 2020 at 17:29 | |||||
| Sep 26, 2020 at 16:25 | comment | added | mathoverflowUser | @JeroenNoels: I checked your claim about $n=945$ and it seems to be true. At least it is not a zero of $f(n,t)$... | |
| Sep 26, 2020 at 16:23 | comment | added | mathoverflowUser | @JeroenNoels: Thanks for your comment. Did you find a counterexample to $f(n,t)$ being irreducible? Thanks for your help. | |
| Sep 26, 2020 at 16:19 | comment | added | Jeroen Noels | That looks very interesting! Thanks for sharing. I did some numerical experiments and found f(n,n) to be negative for n = 945. There could be some patterns here. For example the arithmetic sequence n = 945 + 630 * k provides many more counterexamples (but not all). | |
| S Sep 26, 2020 at 9:19 | history | bounty started | mathoverflowUser | ||
| S Sep 26, 2020 at 9:19 | history | notice added | mathoverflowUser | Canonical answer required | |
| S Sep 25, 2020 at 9:31 | history | suggested | mwt | CC BY-SA 4.0 |
Readability fix
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| Sep 25, 2020 at 7:19 | review | Suggested edits | |||
| S Sep 25, 2020 at 9:31 | |||||
| Sep 25, 2020 at 7:13 | comment | added | mathoverflowUser | also asked on MSE: math.stackexchange.com/questions/3839696/… | |
| Sep 25, 2020 at 6:43 | history | edited | Perfect Number | CC BY-SA 4.0 |
added 847 characters in body; edited tags
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| Sep 24, 2020 at 13:08 | history | edited | Perfect Number | CC BY-SA 4.0 |
added 130 characters in body; edited tags
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| Sep 24, 2020 at 7:23 | history | asked | Perfect Number | CC BY-SA 4.0 |