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
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| Sep 30, 2020 at 4:42 | comment | added | mathoverflowUser | Thanks, I know those formulas. I was hoping to get also a formula for $a_1$. | |
| Sep 29, 2020 at 16:32 | comment | added | Max Alekseyev | @stackExchangeUser: I do not have a formula for $a_1$, but we can prove the following identities: $$a_0 = \frac{A_1}n=\frac{A_0}2,$$ $$a_2 = \frac{A_2-nA_1}2 + na_1.$$ | |
| Sep 29, 2020 at 7:12 | comment | added | mathoverflowUser | It seems it is not possible to derive a formula for $a_i$ using your linear dependence technique, while for $A_i$ it is possible. | |
| Sep 28, 2020 at 16:02 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Sep 28, 2020 at 8:24 | comment | added | mathoverflowUser | ok, I think also that those identities can be proved. | |
| Sep 28, 2020 at 8:24 | comment | added | Max Alekseyev | @stackExchangeUser: Yes, but I'm sure they can be proved. | |
| Sep 28, 2020 at 8:23 | comment | added | mathoverflowUser | So there is no proof for these additional identities, just numerical experiments? | |
| Sep 28, 2020 at 8:22 | comment | added | Max Alekseyev | @stackExchangeUser: See my comment above. | |
| Sep 28, 2020 at 8:20 | comment | added | mathoverflowUser | Could you please provide an indication, how you came up with the last identities? | |
| Sep 28, 2020 at 8:13 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Sep 28, 2020 at 8:06 | vote | accept | mathoverflowUser | ||
| Sep 28, 2020 at 8:06 | comment | added | mathoverflowUser | Yes, That is what I also realised now, that it follows from Ramanujan and van der Pol's identity :) | |
| Sep 28, 2020 at 8:04 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Sep 28, 2020 at 8:00 | comment | added | Max Alekseyev | By finding a linear dependency between vectors of the form $(n^d \sigma_k(n))_{n\geq 1}$ and $(A_2(n))_{n\geq 1}$. | |
| Sep 28, 2020 at 7:54 | comment | added | mathoverflowUser | Thanks, How did you come up with this formula? (Seems that we get nothing new from knowing this in case for perfect numbers...) | |
| Sep 28, 2020 at 7:48 | history | answered | Max Alekseyev | CC BY-SA 4.0 |