Timeline for Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots,x\}.$
Current License: CC BY-SA 4.0
11 events
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| Nov 20, 2021 at 21:29 | comment | added | kodlu | I'd appreciate it if you have a look at the related question mathoverflow.net/questions/408961/… | |
| Nov 20, 2021 at 0:02 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Oct 21, 2020 at 23:01 | comment | added | Alexander Kalmynin | @kodlu I think, it should be $(|A|-o(x))(\ln x)^{a(\ln 2-\delta)}$ (answer edited accordingly), as we first throw away a small bad subset of size $o(x)$ and then evaluate the sum. Also, this does not affect the argument, as $\delta$ is arbitrarily small | |
| Oct 21, 2020 at 22:59 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Oct 20, 2020 at 21:11 | comment | added | kodlu | Looking at this one year later, I cannot seem to justify to myself the steps to fill in the statement: Therefore, for any $A\subset [1,x]\cap \mathbb N$ with $|A|\geq cx$ for some $c>0$ we have $$ S_a^A\geq |A|(\ln x)^{a(\ln 2-\delta)}-o(x), $$ specifically how to obtain the $o(x)$ term. Maybe a step by step sum of numbers with small numbers of divisors, starting with 2, moving on to pseudoprimes, low prime powers, and upper bounding each term? | |
| Sep 18, 2019 at 23:35 | vote | accept | kodlu | ||
| Sep 18, 2019 at 7:37 | comment | added | Alexander Kalmynin | @kodlu, 1. Yes, of course. Corrected, thanks. 2. No, I'm not assuming that $A$ is an interval. I mean that if you discard all the $n$ with large $d(n)$, you will still have enough numbers left to construct your set $A$ | |
| Sep 18, 2019 at 7:34 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Sep 17, 2019 at 23:47 | comment | added | kodlu | And unless I'm totally confused, it should be "if you [discard] all the $n$ with $d(n)>(\ln x)^{\ln 2+\delta}$ out of your set $A$", in the penultimate paragraph. Are you assuming $A$ is an interval in any way? | |
| Sep 17, 2019 at 22:33 | comment | added | kodlu | Thanks. Do you mean "for all but $o(x)$ numbers $n\leq x$ we have $(\ln x)^{\ln 2-\delta}\leq d(n) \leq (\ln x)^{\ln 2+\delta}$? | |
| Sep 17, 2019 at 20:30 | history | answered | Alexander Kalmynin | CC BY-SA 4.0 |