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
Mean of highest exponent in prime factorization of all integers ≥ 2
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number-theory
prime-numbers
prime-factorization
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0@Shreevatsa: I want to know this because i couldn't solve it. By the way have you looked at this problem: http://math.stackexchange.com/questions/1860/comparing-sums-of-reciprocals – 2010-08-11
1 Answers
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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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0Paseman: Could please Tex it out so that the result is more tangible for the viewers. Also i would like you to elaborate it more. – 2010-08-12