Timeline for How are the ratios of successive values of the divisor function distributed?
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6 events
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| Feb 21, 2017 at 17:18 | comment | added | Jan-Christoph Schlage-Puchta | The expectation of $\omega(n)\omega(n+1)$ is $\sum_{p, q: p|n, q|n+1}1$ which is smaller than the expectation of $\omega(n)$ squared, since the latter contains an additional diagonal sum. However, the difference is smaller order than the non-diagonal main term, so the joint distribution of $\omega(n)$ and $\omega(n+1)$ is asymptotically independent with normal marginal distributions (see en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem ). However, for small $x$ the sum of magnitude $\sum_p\frac{1}{p^2}$ is not negligible compared to $\log\log x$, so numerics can be misleading. | |
| Feb 21, 2017 at 15:50 | comment | added | Kevin Smith | Of course I understand that the events $p|n$ and $p|(n+1)$ cannot both happen. But the number of primes less than $n^{1/\loglog n}$ is still large compared to $\log\log n$. | |
| Feb 21, 2017 at 15:37 | comment | added | Kevin Smith | Thank you , Jan-Christoph. Do you mean "$|\log r(n)|<\omega(n)\sqrt{\log\log n}$ for a set of density $1$? Would you clarify what you mean by "negatively correlated" and "do not look independent", please. | |
| Feb 20, 2017 at 21:41 | comment | added | Jan-Christoph Schlage-Puchta | $\log r(n)$ has a normal distribution with mean $2\log\log n$, so if $\omega(n)\rightarrow\infty$, then $|r(n)|<\omega(n)\sqrt{\log\log n}$ for a set of density 1. Note that the events $p|n$ and $p|n+1$ are negatively correlated, so as long as $\log\log n$ is not much larger than 1, $\omega(n)$ and $omega(n+1)$ do not look independent. Also $e^{-\sqrt{\log\log n}/2}$ looks pretty constant for all $n$ accessible by a computer. | |
| Feb 19, 2017 at 21:15 | comment | added | Kevin Smith | When you say $r(n)$ is much closer to $1$, do you know of some estimate that describes how much closer? I'm considering the average value of $e^{-|\log r(n)|/2}$, and my guess is also that this is $>1/2$, perhaps even $\sim \gamma$, based on calculations up to $10^8$. | |
| Feb 13, 2017 at 19:47 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |