Timeline for How many divisors of $n$ are below $n^{1/3}$?
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2021 at 19:59 | comment | added | Will Sawin | Actually, I was wrong about "most". It is not true for all $n$, but a typical $n$ has a large prime factor, meaning that the inequality $d_k(n) \geq d_2(n)/C$ is true with positive probability for all $k$ and all $C \geq 2$, and it might be quite a large probability. | |
| Aug 6, 2021 at 19:04 | comment | added | Will Sawin | $d_3(n)\leq d(n)/2$ is indeed an upper bound, by the obvious symmetry that if $d$ is a divisor of $n$ then $n/d$ is a divisor, and it is attained whenever $n$ is the product of a single large prime $>n^{2/3}$ with some number of smaller primes. However, $d_3(n) \geq d(n)/5$ is certainly not true for all (or even most) $n$, by the concentration of measure statement Terry was mentioning. | |
| Aug 6, 2021 at 17:30 | history | answered | FusRoDah | CC BY-SA 4.0 |