It's a standard result that the number ω(*N*) of prime factors of *N* > 2 can be bounded above by$$ \omega(N) \;=\; \frac{\log(N)}{\log\log(N)} \big(1 + O\big(1/\log\log(N)\big)\big) \;.$$
Are tighter bounds known when *N* is *logarithmically rough* --- that is, where for some fixed constant *c* > 0, *N* has no prime factors smaller than *c* log(*N*)?

The intuition is that $\log_{c\log N}N=\frac{\log N}{\log(c\log N)}=\frac{\log N}{\log c+\log\log N}$ which is asymptotically $\frac{\log N}{\log\log N}$. For this to work, we need that the kth prime for $k=\frac{\log N}{\log(c\log N)}+\pi(c\log N)\approx\frac{\log N}{\log\log N}$ to be 'close' to $c\log N$ -- actually, any constant multiple of $\log N$ will do.

$p_k\approx\frac{\log N}{\log\log N}\log\frac{\log N}{\log\log N}\approx\log N$, so $\log_{p_k}N\approx\frac{\log N}{\log\log N}$, as desired. This could probably made explicit with Rosser's theorem and/or Dusart's various bounds on $p_n$ and $\pi(n)$.

(a)how do you obtain the bound $y\sim(\frac{c}{c+1})x$, and(b)how do you obtain the asymptotic growth of $N$ with respect to $x$? For the latter (given the ratio of $y$ to $x$ described) I obtain a rough bound of $N>\frac{x!}{[cx/(c+1)]!}\in\Omega\big((x/e)^{x/(c+1)}\big)$ , which is larger than the growth you describe. Could you elaborate further on your scribblings, or refer me to the (likely elementary textbook) reference where I could find the tools you're using? $\endgroup$ – Niel de Beaudrap May 10 '10 at 7:351more comment