# What is the relative size of the radical of an ABC-triple relative to the number of primes up to its largest element?

Write $\bf N$ for the set of natural numbers, and $P$ for the set of primes. For $x$ in $\bf N$ let $p(x)$ be the product of the primes dividing $x$ (that is, the "radical" of $x$). Also write $\#(x)$ for the number of primes up to $x$. Let $S$ be the set of triples $(x,a,b)$ of naturals with $\gcd(a,b)=1$ and $x=a+b$. For $s=(x,a,b)$ in $S$, let $f(s)=\#(x)/(p(x)p(a)p(b))$. What is $\limsup_{s{\rm\ in\ }S}f(s)$?

In more classical notation you seem to ask for $\frac{\pi(c)}{\operatorname{rad}(abc)}$ for an ABC-triple $c= a+ b$.

By results of van Frankenhuysen one has infinitely many triples such that $$\log c \ge \log \operatorname{rad}(abc) + k \sqrt{\log c / \log \log c}$$ for some positive $k$.

Written differently $$\frac{c}{\operatorname{rad}(abc)} \ge \exp( k \sqrt{\log c / \log \log c} )$$

Now $\pi(c) \sim c/ \log c$, so for sufficiently large $c$

$$\frac{\pi(c)}{\operatorname{rad}(abc)} \ge k'\frac{\exp( k \sqrt{\log c / \log \log c} )}{\log c}$$ for some positive $k'$.

The right-hand side tends to infinity as $c$ tends to infinity, and so the limsup you ask about is infinite.

Machiel van Frankenhuysen, A lower bound in the $abc$ conjecture, J. Number Theory 82 (2000), no. 1, 91--95.