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Re: [PrimeNumbers] Number of factors in average

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  • mikeoakes2@aol.com
    In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@terra.es ... It tends to infinity. The standard number theory function you are interested in
    Message 1 of 6 , May 5, 2003
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      In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@...
      writes:


      > Which is the best manner to calculate the average of factors of the
      > numbers up to a fixed N? And between N and M?
      >
      > (for example, up to 10: 1,2...,9,10 have 1,1,1,2,1,2,1,3,2,2 factors, the
      > average is 1,6)
      >
      > Is the average tending to infinite or to a specific ratio?
      >

      It tends to infinity.

      The standard number theory function you are interested in is d(n), defined to
      be the number of divisors of n, including 1 and n, so that d(n) >= 2.

      Then Hardy and Wright "An Introduction to the Theory of Numbers" (1979) (p.
      264) have :-
      "Theorem 319. The average order of d(n) is log n."

      An alternative formulation of this result is:-
      lim {n -> oo} [d(1) + d(2) + ... + d(n)] / log(n) = 1.

      Mike Oakes


      [Non-text portions of this message have been removed]
    • mikeoakes2@aol.com
      Sent too quickly - I should have written: lim {n - oo} [d(1) + d(2) + ... + d(n)] / (n*log(n)) = 1. Mike [Non-text portions of this message have been removed]
      Message 2 of 6 , May 5, 2003
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        Sent too quickly - I should have written:
        lim {n -> oo} [d(1) + d(2) + ... + d(n)] / (n*log(n)) = 1.

        Mike


        [Non-text portions of this message have been removed]
      • Jose Ramón Brox
        Hi: 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? Jose ... From:
        Message 3 of 6 , May 5, 2003
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          Hi:

          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?

          Jose
          ----- Original Message -----
          From: mikeoakes2@...
          To: primenumbers@yahoogroups.com
          Sent: Monday, May 05, 2003 1:39 PM
          Subject: Re: [PrimeNumbers] Number of factors in average


          In a message dated 05/05/03 10:25:14 GMT Daylight Time, ambroxius@...
          writes:


          > Which is the best manner to calculate the average of factors of the
          > numbers up to a fixed N? And between N and M?
          >
          > (for example, up to 10: 1,2...,9,10 have 1,1,1,2,1,2,1,3,2,2 factors, the
          > average is 1,6)
          >
          > Is the average tending to infinite or to a specific ratio?
          >

          It tends to infinity.

          The standard number theory function you are interested in is d(n), defined to
          be the number of divisors of n, including 1 and n, so that d(n) >= 2.

          Then Hardy and Wright "An Introduction to the Theory of Numbers" (1979) (p.
          264) have :-
          "Theorem 319. The average order of d(n) is log n."

          An alternative formulation of this result is:-
          lim {n -> oo} [d(1) + d(2) + ... + d(n)] / log(n) = 1.

          Mike Oakes


          [Non-text portions of this message have been removed]


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          [Non-text portions of this message have been removed]
        • mikeoakes2@aol.com
          In a message dated 06/05/03 00:40:56 GMT Daylight Time, ambroxius@terra.es ... them is quite ... Sorry, my fault. That standard number theory function is
          Message 4 of 6 , May 5, 2003
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            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]
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