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
6 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2019 at 23:35 | vote | accept | kodlu | ||
| Sep 17, 2019 at 20:30 | answer | added | Alexander Kalmynin | timeline score: 4 | |
| Sep 17, 2019 at 18:10 | comment | added | Greg Martin | The quantity $c\log x/\log\log x$ can't possibly affect the size of the lower bound, since the contribution of $d(n)^a$ over that few integers is $\ll x^\varepsilon$. I imagine that the answer will be essentially $x(\log x)^{a\log 2}$ (for $a>0$), since as you say most integers have around $2^{\log\log n} = (\log n)^{\log 2}$ divisors. (That the exponent of $\log x$ in the asymptotic for $S_a(x)$ grows exponentially with $a$ is due to a relatively few integers with a huge number of divisors.) | |
| Sep 17, 2019 at 16:51 | answer | added | Dr. Pi | timeline score: 1 | |
| Sep 16, 2019 at 2:07 | history | edited | LeechLattice |
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| Sep 16, 2019 at 1:11 | history | asked | kodlu | CC BY-SA 4.0 |