Timeline for Is $\sum_{n=1}^\infty\frac{S(n)}{n!}$ an irrational, where $S(n)$ denotes the sum of remainders function?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2019 at 20:59 | comment | added | Sylvain JULIEN | The number of digits of the n-th term seems to be about n/2. | |
| Sep 7, 2019 at 11:22 | comment | added | user142929 | Many thanks for share your great calculations, I am going to study these. | |
| Sep 6, 2019 at 23:10 | comment | added | Thomas Browning | @IlyaBogdanov it looks infinite to me but I don't see how one would prove it | |
| Sep 6, 2019 at 23:09 | comment | added | Thomas Browning | Here's the OEIS sequence: A056550 | |
| Sep 6, 2019 at 22:57 | comment | added | Ilya Bogdanov | @ThomasBrowning: I've obtained this formula as well. Do you have any idea of how to search those attempting this question? | |
| Sep 6, 2019 at 22:53 | comment | added | Thomas Browning | $S(n)=n^2-\sum_1^n\sigma(k)$ where $\sigma$ is the sum of divisors function. Surely someone has thought about whether $n$ divides $\sum_1^n\sigma(n)$ infinitely often. | |
| Sep 6, 2019 at 22:42 | comment | added | Fedor Petrov | It seems to help, if we accurately specify the remainder term. Say, take $n$ which is simultaneously divisible by $100!$ and such that $n\beta$ is almost integer, where $\beta=1-\pi^2/12$ is the limit value. | |
| Sep 6, 2019 at 22:03 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |