Loading ...
Sorry, an error occurred while loading the content.

12363Re: [PrimeNumbers] Number of factors in average

Expand Messages
  • mikeoakes2@aol.com
    May 5 11:44 PM
    • 0 Attachment
      In a message dated 06/05/03 00:40:56 GMT Daylight Time, ambroxius@...
      writes:

      > I was thinking on prime factors rather than in divisors... the number of
      them is quite
      > lesser than this of the divisors... what order has?

      Sorry, my fault.
      That standard number theory function is Omega(n), defined to be the total
      number of prime factors of n; in other words, if there is the prime
      factorisation
      n = p_1^e_1 * ... * p_r^e_r,
      then
      Omega(n) = e_1 + ... + e_r.

      So, in particular Omega(1) = 0. [As an aside: anyone who thinks 1 is a prime
      would have a hard job defining Omega(); and 1 is certainly not composite...]

      Omega(n) has average order log(log(n)).

      Mike


      [Non-text portions of this message have been removed]
    • Show all 6 messages in this topic