Timeline for A truncated divisor sum
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2019 at 0:51 | vote | accept | user164144 | ||
| Jan 16, 2019 at 0:51 | comment | added | user164144 | I get it now. he latter bound you mention of $A^{-2+o(1)}$ is one that I can show, but is not useful to me. Thanks though. | |
| Jan 14, 2019 at 23:26 | comment | added | Aleksei Kulikov | @user164144 Sorry, I should have written it more clearly. Of course I didn't provide lower bound anywhere close to $d(N)/A^3$(and I doubt that this is possible), but I gave an example with $\exp(C\log(N)/\log\log(N))/A^3$ which differs from $d(N)/A^3$ only by constant $C$ in exponent. Yet it is more than enough to prove that nothing close to $\log(N)^m/A^3$ is possible. I also have similar example for small $A$ which gives something like $1/A^{2-o(1)}$, I can add it if it is of interest for you. | |
| Jan 14, 2019 at 16:38 | comment | added | user164144 | Aleksei, you have shown above that the number of divisors of $n$ in (A,2A] can be $\text{exp}(O(\log(n)/\log\log(n)))$. Why that implies what $\sum_{\substack{d|N \\ d>A}}d^{-3}$ has an optimal upper bound of $d(N)/A^3$? For instance, $\sum_{d|n}d^{-1}$ has an upper bound $\log\log(n)$ which is smaller than an upper bound for $d(n).$ | |
| Jan 13, 2019 at 23:17 | history | answered | Aleksei Kulikov | CC BY-SA 4.0 |