Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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

share|cite|improve this question
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. – kennytm Aug 11 '10 at 9:47
@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 – anonymous Aug 11 '10 at 14:39
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. – J. M. Aug 11 '10 at 14:46
@J.Mangaldan:- I agree. – anonymous Aug 11 '10 at 15:27
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. – ShreevatsaR Aug 11 '10 at 15:45
up vote 1 down vote accepted

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.

share|cite|improve this answer
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. – anonymous Aug 12 '10 at 13:44

Your Answer


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