Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

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.

share|cite|improve this answer
+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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