0
$\begingroup$

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 $\limsup_{s{\rm\ in\ }S}f(s)$?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.