For any natural number $n > 1$, define $E(n)$,to be the highest exponent to which a prime divides it. For instance, $E(12)=E(36)=2$. Show that $$\lim_{N \to \infty} \frac{1}{N} \sum\limits_{n=2}^{N} E(n)$$ exists and find its value
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3$\begingroup$ Can you give a title that is more specific next time you ask a question? The title is used for searching so it's important be precise. $\endgroup$– kennytmCommented Aug 11, 2010 at 9:47
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1$\begingroup$ @Kenny TM: hi, i can only give the title which comes to my mind at that time. In case you don't like it then please be free to edit it. Sorry $\endgroup$– anonymousCommented Aug 11, 2010 at 14:39
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5$\begingroup$ That sounds very sloppy. Coming up with a useful title is one of the few tokens of respect you can give to would-be answerers. $\endgroup$– J. M. ain't a mathematicianCommented Aug 11, 2010 at 14:46
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$\begingroup$ @J.Mangaldan:- I agree. $\endgroup$– anonymousCommented Aug 11, 2010 at 15:27
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2$\begingroup$ As with all your questions: (1) where is this question from, (2) why you want to know? If you provide more personal context and motivation, I suspect more people will feel like trying/answering. $\endgroup$– ShreevatsaRCommented Aug 11, 2010 at 15:45
1 Answer
Here are some suggestions to approximate the limit.
Consider F(n) to be the greater of 1 and the highest power of 2 that appears in the prime factorization of n. Show what the limit of the average over n of F(n) is as n gets large.
Let G_p(n) be defined like F(n), except replace 2 by an arbitrary prime p.
Note that F(n) <= E(n), and that E(n) = max(G_p(n) over all primes p), so that if limit of the average over n of E(n) exists, you can bound it using a sum of limits involving G_p.
For more accurate estimates, consider inclusion-exclusion.
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$\begingroup$ Paseman: Could please Tex it out so that the result is more tangible for the viewers. Also i would like you to elaborate it more. $\endgroup$– anonymousCommented Aug 12, 2010 at 13:44