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Nov 15, 2015 at 21:15 comment added Serguei Popov Now I think that the other answer ($\Omega(n/\sqrt{\log n})$) may be true, but I'm pretty sure it should be $o(n)$ in any case. Well, just intuitive feeling :)
Nov 15, 2015 at 19:50 comment added Anthony Quas I also thought about this heuristic, but I feel it might not be right. It's true that you delete on average $\log n$ numbers in your first step, but (1) much of this is accounted for by very small divisors - which can only be used once; and (2) once you've used up a small divisor, later removals kill off a lot less. My money is on $\Omega(n)$. (Also in the computer experiment, you see a lot of time where "nothing" is being deleted and $\sigma(n)/n$ decreases from $0.28$ when $n=1000$ to $0.24$ when $n=1000000$.)
Nov 15, 2015 at 14:48 history answered Serguei Popov CC BY-SA 3.0