Timeline for Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 22 at 12:36 | comment | added | mathoverflowUser | @TatendaKubalalika thanks | |
| Aug 22 at 12:24 | comment | added | Q_p | Yes. I believe your equation is quite interesting, though. | |
| Aug 22 at 12:16 | comment | added | mathoverflowUser | @TatendaKubalalika: The error was in the sign of $3\sigma_3(n)n^2$ which led you to conclude that $0 \equiv 6n^3-2n^3 \equiv 4n^3 \mod(8)$ while it should have been $0 \equiv 6n^3+2n^3 \mod(8)$ which evaluates to $0 \mod (8)$ without giving any hint if $n$ is even or odd. | |
| Aug 22 at 12:12 | history | edited | mathoverflowUser | CC BY-SA 4.0 |
corrected formula
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| Aug 22 at 11:45 | comment | added | Q_p | Great, so i'll delete my above comments as they are now obsolete. | |
| Aug 22 at 11:37 | comment | added | mathoverflowUser | @TatendaKubalalika: I have checked the equation satisfied by the odd perfect number and it seems that this equation must be: $0=-120n^4+45n^2 \sigma_3(n)+30n^3-360A_2$, which evaluated with your method gives $0 \equiv 96n^3 \mod(8)$ , so one can not conclude that $n$ must be even... I will update the question with the correct equation. | |
| Aug 22 at 10:36 | comment | added | mathoverflowUser | @Q_p: Ok, I will post a question here, with remark to your comment. | |
| Aug 22 at 10:33 | comment | added | mathoverflowUser | @Q_p: Strange, indeed! I would post a question here asking for clarification, if there is an error in my equation or ... :-) . It has been 3 years ago, and I am not sure if I have the notebook with the calculation to verify it. | |
| Aug 22 at 10:31 | comment | added | Q_p | You have $8n^4 - 2n^3 + 3\sigma_{3}(n)n^2 + 24A_2=0$ thus $-2n^3 + 3\sigma_{3}(n)n^2$ must be divisible by $8$. Since $\sigma_{3}(n) \equiv \sigma(n)=2n \mod 8$, it follows that $-2n^3 + 6n^3 = 4n^3$ must be divisible by $8$. | |
| Aug 22 at 10:29 | comment | added | mathoverflowUser | @Q_p: Thanks for your comment. Interesting thougt. I had not thought about it in this direction. But why is $4n^3 \equiv 0 \mod 8$? | |
| Aug 22 at 10:25 | comment | added | Q_p | Are you sure you didn't make some arithmetical error when you substituted $\sigma(n)=2n$ into van der Pol's equation and then applied Ramanujan's identity? I mean, one can easily deduce from your equation $8n^4 - 2n^3 + 3\sigma_{3}(n)n^2 + 24A_2=0$ that every perfect number must be even. Indeed, simply consider that equation $\mod 8$, and note that $d^3 \equiv d$ mod $8$ if $d$ is odd thus $\sigma_{3}(n) \equiv \sigma(n)=2n$ mod $8$. This entails that $8n^4 - 2n^3 + 3\sigma_{3}(n)n^2 + 24A_2 \equiv 6n^3 -2n^3 = 4n^3 \equiv 0$ mod $8$, thus $n$ must be even. | |
| Sep 28, 2020 at 11:31 | answer | added | Henri Cohen | timeline score: 9 | |
| Sep 28, 2020 at 8:06 | vote | accept | mathoverflowUser | ||
| Sep 28, 2020 at 7:48 | answer | added | Max Alekseyev | timeline score: 9 | |
| Sep 28, 2020 at 6:02 | review | First posts | |||
| Sep 28, 2020 at 6:28 | |||||
| Sep 28, 2020 at 5:57 | history | asked | mathoverflowUser | CC BY-SA 4.0 |