Timeline for What is known about the mode of the number of divisors $\le x$?
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| Sep 25, 2019 at 23:13 | comment | added | MyNinthAccount | For the related question of the mode of $\Omega(n)$: can you use Sathe/Selberg for small $k$, then Nicolas for large $k$, and maybe Delange in the transition range? doi.org/10.1006/jnth.1997.2216 | |
| Sep 25, 2019 at 23:06 | comment | added | Gerry Myerson | A similar question was raised at m.se last year, math.stackexchange.com/questions/2774937/… | |
| Sep 25, 2019 at 22:43 | history | edited | user514014 | CC BY-SA 4.0 |
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| Sep 25, 2019 at 22:27 | comment | added | user514014 | Thanks, I should've caught that! | |
| Sep 25, 2019 at 22:25 | comment | added | Greg Martin | Rephrasing Stanley's comment in a streamlined form: the divergence of $M(x)$ follows from the two statements $d(n)\ge 2^{\omega(n)}$ (where $\omega(n)$ is the number of distinct prime factors of $n$) and "For any fixed $k$, the set of integers $n$ with $\omega(n)\le k$ has density $0$." | |
| Sep 25, 2019 at 22:21 | comment | added | Stanley Yao Xiao | It's a much harder question to estimate $M(x)$ precisely (i.e., an asymptotic relation), since one would need to deal with cases where uniformity estimates are not available. | |
| Sep 25, 2019 at 22:21 | comment | added | Stanley Yao Xiao | $M(x)$ will go to infinity as $x$ tends to infinity. Fixing $k \geq 1$, one can choose $x$ sufficiently large so that the number of numbers in $[1,x]$ with exactly $k$ prime factors is smaller than those with $k+1$ prime factors, by the result of Landau you mentioned. Indeed, Montgomery and Vaughan gives quite precise estimates and ranges of uniformity for the asymptotic formula. This shows that no fixed $k$ can be the mode for all $x$. | |
| Sep 25, 2019 at 21:58 | history | asked | user514014 | CC BY-SA 4.0 |