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For integer $n>1$ define $q(n)=\frac{\log(\rm{rad}(n))}{\log(n)}$ where $\rm{rad}(n)$ is the radical of $n$, the product of the disctinct prime factors.

For real $A$ and integer $N$ define $S_{N,A}=\#\{n : 1 < n <N,q(n)\le A\}$.

$s(N,A)=\frac{S_{N,A}}{N}$ and $s^*(N,A)=\frac{\log(S_{N,A})}{\log(N)}$.

Q1. Are there lower bounds for $s(N,A)$ or $s^*(N,A)$?

Q2. Is $s^*(N,1/2) \ge 1/2 +C$ for sufficiently large $N$ and fixed positive $C$?

$1/2$ is trivial lower bound because of squares. Limited numerical evidence up to $10^7$ suggests that $s^*(N,1/2) = 0.62\ldots$.

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It may be more natural to consider $$ N(x,y) = \# \{ n\le x: \text{rad}(n) \le y\}, $$ and the question is essentially about $N(x,x^{\alpha})$ (with $\alpha =1/2$ in your question 2). The quantity $N(x,y)$ has been analyzed in detail by Robert and Tenenbaum, and this analysis has been used by Robert, Stewart and Tenenbaum to formulate a refined version of the abc-conjecture. In the range $y \ge \exp ((\log x)^{\frac 12+\epsilon})$ one has $$ N(x,y) \sim y F(v), $$ where $v= \log (x/y)$, and $$ F(v) = \exp\Big( \sqrt{\frac{8v}{\log v}} (1+o(1))\Big). $$ (These are rough versions of more precise results from the cited papers.) In particular one has $$ N(x,\sqrt{x}) \sim \sqrt{x} \exp \Big( \sqrt{\frac{4\log x}{\log \log x}}(1+o(1))\Big). $$ So the limited numerical evidence in question 2 is misleading, and the right exponent there is just $1/2$.

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  • $\begingroup$ Thank you. Is $s^*(N,A)=A$ for $0 < A < 1$? $\endgroup$
    – joro
    Commented Apr 2, 2018 at 8:47
  • $\begingroup$ Yes (as $N\to \infty$). $\endgroup$
    – Lucia
    Commented Apr 2, 2018 at 13:31

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