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

Number of factors in average

Expand Messages
  • Jose Ramón Brox
    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
    Message 1 of 6 , May 5, 2003
    • 0 Attachment
      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?

      Jose Brox
      http://espanol.groups.yahoo.com/group/Telecomunicacion/





      [Non-text portions of this message have been removed]
    • Décio Luiz Gazzoni Filho
      ... Hash: SHA1 ... If no one else has a better approach, I suggest a statistical approach based on Dickman s theorem on smoothness of number, i.e. psi(n,
      Message 2 of 6 , May 5, 2003
      • 0 Attachment
        -----BEGIN PGP SIGNED MESSAGE-----
        Hash: SHA1

        On Monday 05 May 2003 06:24, Jose Ramón Brox wrote:
        > 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?
        >

        If no one else has a better approach, I suggest a statistical approach based
        on Dickman's theorem on smoothness of number, i.e. psi(n, n^(1/x)) = x*u^(-u)
        with u = log(x)/log(log(x)). Now compute the expected value E[ ] (you should
        be very familiar with that given your telecommunications background!) and
        luckily you'll be able to work out a formula.

        Décio
        -----BEGIN PGP SIGNATURE-----
        Version: GnuPG v1.2.1 (GNU/Linux)

        iD8DBQE+tkI5ce3VljctsGsRAmyzAKDEu1uKK9mmAXOc7chpZluQ0AFb+ACfWjwV
        8pjLE6eyH0wdMqphM7Xe69U=
        =as3k
        -----END PGP SIGNATURE-----
      • 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 3 of 6 , May 5, 2003
        • 0 Attachment
          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 4 of 6 , May 5, 2003
          • 0 Attachment
            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 5 of 6 , May 5, 2003
            • 0 Attachment
              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]


              Yahoo! Groups Sponsor





              Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
              The Prime Pages : http://www.primepages.org/



              Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.



              [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 6 of 6 , May 5, 2003
              • 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]
              Your message has been successfully submitted and would be delivered to recipients shortly.