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Question: What interesting or notable problems involve the additive function $\sum_{p \mid n} \frac{1}{p}$ and depend on sharp bounds of this function?

I'm aware of at least one: A certain upper bound of the sum implies the ABC conjecture.

By the Arithmetic-Geometric Inequality, one can bound the radical function (squarefree kernel) of an integer, \begin{align} \left( \frac{\omega(n)}{\sum_{p \mid n} \frac{1}{p}} \right)^{\omega(n)} \leqslant \text{rad}(n). \end{align} If for any $\epsilon > 0$ there exists a finite constant $K_{\epsilon}$ such that for any triple $(a,b,c)$ of coprime positive integers, where $c = a + b$, one has \begin{align} \sum_{p \mid abc} \frac{1}{p} < \omega(abc) \left( \frac{K_\epsilon}{c} \right)^{1/((1+\epsilon) \omega(abc))}, \end{align} then \begin{align} c < K_{\epsilon} \left( \frac{\omega(abc)}{\sum_{p \mid abc} \frac{1}{p}} \right)^{(1+ \epsilon)\omega(abc)} \leqslant \text{rad}(abc)^{1+\epsilon}, \end{align} and the ABC-conjecture is true.

Edit: Now, whether or not any triples satisfy the bound on inverse primes is a separate issue. Greg Martin points out that there are infinitely many triples which indeed violate it. This begs the question of whether there are any further refinements of the arithmetic-geometric inequality which remove such anomalies, but this question is secondary.

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I've removed literature-search tag; reference-request already says this. –  Martin Sleziak Aug 2 '12 at 15:05

1 Answer 1

As it turns out, your proposed inequality is false - in fact false for any $\epsilon>0$, even huge $\epsilon$.

The following approximation to the twin primes conjecture was proved by Chen's method: there exist infinitely many primes $b$ such that $c=b+2$ is either prime or the product of two primes. With $a=2$, this gives $\omega(abc)\le4$, and so $$ \omega(abc) \left( \frac{K_\epsilon}{c} \right)^{1/((1+\epsilon) \omega(abc))} \le 4 \bigg( \frac{K_\epsilon}c \bigg)^{1/(4+4\epsilon)}, $$ which (for any fixed $\epsilon>0$) can be arbitrarily small as $c$ increases. However, the other side $\sum_{p\mid abc} \frac1p$ is at least $\frac12$, and so the inequality has infinitely many counterexamples.

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+1 This is good to know, but not really the point. Might you have any suggestions for interesting or notable problems where the sum in question is essential? Thanks! –  user02138 Aug 2 '12 at 21:45

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