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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.

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The symbol you wanted there is $\gg$, produced by \gg. –  joriki Jul 22 '12 at 12:55
    
Thanks, edited. –  daniel Jul 22 '12 at 13:03

1 Answer 1

up vote 7 down vote accepted

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

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1  
I don't know whether to infer from the counterexample a negative response to the broader question. Will wait a bit before accepting this. –  daniel Jul 22 '12 at 13:42
2  
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? –  Gerry Myerson Jul 22 '12 at 23:38
    
Yes, am doing so. –  daniel Jul 23 '12 at 20:51

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