Let a-f be integers g.t. 2 with $a < b < c < d < e < f$. Let

$$\ln def - \ln a b c = \alpha.$$

Let $\{p_i\}$ be the set of prime factors (with repetitions) in a,b,c. Let $\{q_i\}$ be the set of prime factors in d,e,f. Then we know that

$$ \ln def = \ln \prod q_i = \sum \ln q_i$$ and so for $p_i$ and a,b,c.

Then $$\sum \ln q_i - \sum \ln p_i = \alpha .$$

It is true I think that $def - abc \gg \ln def - \ln abc.$ My question is, can we make any quantitative statements $ \alpha = f (\beta)$ about

$$\sum q_i - \sum p_i = \beta$$ based on our knowledge of $\alpha$? It's tempting to say that $\alpha < \beta$, for example, but I don't see how to prove it.

Thanks for any suggestions/answers.

  • $\begingroup$ The symbol you wanted there is $\gg$, produced by \gg. $\endgroup$ – joriki Jul 22 '12 at 12:55

$$17\lt19\lt23\lt25\lt27\lt32$$ but $$17+19+23\gt5+5+3+3+3+2+2+2+2+2$$ so $\alpha\gt0$ but $\beta\lt0$.

  • 1
    $\begingroup$ I don't know whether to infer from the counterexample a negative response to the broader question. Will wait a bit before accepting this. $\endgroup$ – daniel Jul 22 '12 at 13:42
  • 2
    $\begingroup$ I suspect the truth is that given any positive $r$, any real $s$, and any positive $\epsilon$ there are $a,b,c,d,e,f$ meeting your conditions with $|r-\alpha|\lt\epsilon$ and $|s-\beta|\lt\epsilon$; this would say that information about $\alpha$ tells you nothing about $\beta$. Why not play around with it, construct a few examples, get some feel for the problem? $\endgroup$ – Gerry Myerson Jul 22 '12 at 23:38
  • $\begingroup$ Yes, am doing so. $\endgroup$ – daniel Jul 23 '12 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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