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12363Re: [PrimeNumbers] Number of factors in average

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  • mikeoakes2@aol.com
    May 5, 2003
      In a message dated 06/05/03 00:40:56 GMT Daylight Time, ambroxius@...

      > 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
      n = p_1^e_1 * ... * p_r^e_r,
      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)).


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