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Re: [PrimeNumbers] Number of factors in average

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  • 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 1 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 2 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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      • 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 3 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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