Timeline for Sum of divisors below threshold
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2018 at 9:47 | vote | accept | Kurisuto Asutora | ||
| Jun 16, 2018 at 3:23 | history | edited | GH from MO |
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| Jun 16, 2018 at 0:23 | answer | added | Lucia | timeline score: 11 | |
| Jun 15, 2018 at 7:52 | comment | added | Greg Martin | Try using Rankin's trick: $\sum_{d\mid n,\, d\le D} d \le \sum_{d\mid n} d (D/d)^\epsilon$, the right-hand side of which is a multiplicative function of $n$. I think that taking $\epsilon = o(1/\log\log n)$ yields the result that $D = n/\exp\big( (\log\log n)^2 f(n) \big)$ is valid if $f(n)\to\infty$. | |
| Jun 15, 2018 at 7:38 | comment | added | Greg Martin | When $n$ is so highly composite that $\sigma(n) > (e^\gamma-o(1))n\log\log n$, then the divisors above $D$ have to include enough numbers in the set $n,\frac n2, \frac n3,\dots,\frac n{n/D}$ so that their sum is at least $(e^\gamma-o(1))n\log\log n$. This means that $D$ cannot be taken to be $n/(\log n)^{e^\gamma-\epsilon}$ for any $\epsilon>0$. | |
| Jun 15, 2018 at 7:11 | history | edited | Martin Sleziak |
The tag (divisors) is for divisors in algebraic geometry - see the tag-info https://mathoverflow.net/tags/divisors/info
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| Jun 15, 2018 at 4:03 | comment | added | Kurisuto Asutora | Yes, of course one may find bounds on D assuming additional number-theoretic properties of n, such as having only large prime powers. But this is not what I want to have, D is supposed to be a "nice" function of n which does not consider arithmetic properties but depends only on the size of n. | |
| Jun 14, 2018 at 16:02 | comment | added | Gerhard Paseman | Also, the "right" order for n may be wrong if all (or enough) of the prime factors of n are bigger than log n. Gerhard "Talking About Really Big N" Paseman, 2018.06.14. | |
| Jun 14, 2018 at 15:55 | comment | added | Gerhard Paseman | You might consider the partial sum as a function S(D,n), and play around with inequalities. For example, when is (1+p)S(D,n) greater than S(pD,pn)? You may find that you can tweak D in a multiplicative fashion to get some bound that works for you. Gerhard "Play In A Bigger Field" Paseman, 2018.06.14. | |
| Jun 14, 2018 at 1:58 | history | asked | Kurisuto Asutora | CC BY-SA 4.0 |