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Divisor Function

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  • Sebastian Martin
    Hello all: I have obtained this result: DivisorSigma[2k,m]/DivisorSigma[k,m] is integer if and only if m is a perfect square DivisorSigma[k,m]=sum[d^k, d
    Message 1 of 7 , Aug 19 4:20 AM
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      Hello all:

      I have obtained this result:

      DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
      if and only if
      m is a perfect square

      DivisorSigma[k,m]=sum[d^k, d divisor of m]

      Can anyone prove this result?

      Sincerely

      Sebastian Martin Ruiz


      ---------------------------------

      Sé un Mejor Amante del Cine
      ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.


      [Non-text portions of this message have been removed]
    • Adam
      ... I obtain a lot of small counterexamples with k=1, some with k=2, and ... 4 (3) (7) ... q1 := 2726235765168410 q2 := 7247255655544865674860411008810
      Message 2 of 7 , Aug 21 10:35 AM
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        --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
        wrote:
        >
        > Hello all:
        >
        > I have obtained this result:
        >
        > DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
        > if and only if
        > m is a perfect square
        >
        > DivisorSigma[k,m]=sum[d^k, d divisor of m]
        >
        > Can anyone prove this result?
        >
        > Sincerely
        >
        > Sebastian Martin Ruiz
        >
        >
        > ---------------------------------
        >
        > Sé un Mejor Amante del Cine
        > ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
        >
        >
        > [Non-text portions of this message have been removed]
        >

        I obtain a lot of small counterexamples with k=1, some with k=2, and
        k=3,m=6050 and k=4,m=7203. E.g., Maple output:

        > ifactor(7203);
        4
        (3) (7)
        > q1:=sigma[4](7203);
        > q2:=sigma[8](7203);
        > q2/q1;
        q1 := 2726235765168410
        q2 := 7247255655544865674860411008810
        2658337825414441


        All of my counterexamples "appear" to have square factors tho'.....
      • Sebastian Martin
        7203 is NOT a perfect square The result is true only if only m is a perfect square: m=n^2 m= 1, 4, 9, 16, 25, 36, ...... ... I obtain a lot of small
        Message 3 of 7 , Aug 21 1:00 PM
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          7203 is NOT a perfect square

          The result is true only if only m is a perfect square: m=n^2

          m= 1, 4, 9, 16, 25, 36, ......

          Adam <a_math_guy@...> escribió:
          --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
          wrote:
          >
          > Hello all:
          >
          > I have obtained this result:
          >
          > DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
          > if and only if
          > m is a perfect square
          >
          > DivisorSigma[k,m]=sum[d^k, d divisor of m]
          >
          > Can anyone prove this result?
          >
          > Sincerely
          >
          > Sebastian Martin Ruiz
          >
          >
          > ---------------------------------
          >
          > Sé un Mejor Amante del Cine
          > ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
          >
          >
          > [Non-text portions of this message have been removed]
          >

          I obtain a lot of small counterexamples with k=1, some with k=2, and
          k=3,m=6050 and k=4,m=7203. E.g., Maple output:

          > ifactor(7203);
          4
          (3) (7)
          > q1:=sigma[4](7203);
          > q2:=sigma[8](7203);
          > q2/q1;
          q1 := 2726235765168410
          q2 := 7247255655544865674860411008810
          2658337825414441

          All of my counterexamples "appear" to have square factors tho'.....






          ---------------------------------

          ¡Descubre una nueva forma de obtener respuestas a tus preguntas!
          Entra en Yahoo! Respuestas.


          [Non-text portions of this message have been removed]
        • Adam
          7203 is not a perfect square! Maybe you didn t mean if and only if. Part of if and only if is: sigma[2k](m)/sigma[k](m) is integer only if m is a square I
          Message 4 of 7 , Aug 21 1:10 PM
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            7203 is not a perfect square!

            Maybe you didn't mean "if and only if." Part of "if and only if" is:

            sigma[2k](m)/sigma[k](m) is integer only if m is a square

            I demonstrated by example: there exists a non square for which the
            fraction is an integer.

            --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
            wrote:
            >
            > 7203 is NOT a perfect square
            >
            > The result is true only if only m is a perfect square: m=n^2
            >
            > m= 1, 4, 9, 16, 25, 36, ......
            >
            > Adam <a_math_guy@...> escribió:
            > --- In primenumbers@yahoogroups.com, Sebastian Martin
            <sebi_sebi@>
            > wrote:
            > >
            > > Hello all:
            > >
            > > I have obtained this result:
            > >
            > > DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
            > > if and only if
            > > m is a perfect square
            > >
            > > DivisorSigma[k,m]=sum[d^k, d divisor of m]
            > >
            > > Can anyone prove this result?
            > >
            > > Sincerely
            > >
            > > Sebastian Martin Ruiz
            > >
            > >
            > > ---------------------------------
            > >
            > > Sé un Mejor Amante del Cine
            > > ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
            > >
            > >
            > > [Non-text portions of this message have been removed]
            > >
            >
            > I obtain a lot of small counterexamples with k=1, some with k=2,
            and
            > k=3,m=6050 and k=4,m=7203. E.g., Maple output:
            >
            > > ifactor(7203);
            > 4
            > (3) (7)
            > > q1:=sigma[4](7203);
            > > q2:=sigma[8](7203);
            > > q2/q1;
            > q1 := 2726235765168410
            > q2 := 7247255655544865674860411008810
            > 2658337825414441
            >
            > All of my counterexamples "appear" to have square factors tho'.....
            >
            >
            >
            >
            >
            >
            > ---------------------------------
            >
            > ¡Descubre una nueva forma de obtener respuestas a tus preguntas!
            > Entra en Yahoo! Respuestas.
            >
            >
            > [Non-text portions of this message have been removed]
            >
          • Peter Kosinar
            ... As Adam already disproved the only if part, let s concentrate on the if one. First, there are two simple observations to make: 1) If p is a prime,
            Message 5 of 7 , Aug 21 2:04 PM
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              > DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
              > if and only if
              > m is a perfect square
              >
              > DivisorSigma[k,m]=sum[d^k, d divisor of m]

              As Adam already disproved the "only if" part, let's concentrate on the
              "if" one. First, there are two simple observations to make:
              1) If p is a prime,
              DivisorSigma[k, p^e] = 1 + p^k + ... + p^ek = (p^[k(e+1)] - 1)/(p^k-1)
              2) If c and d are coprime,
              DivisorSigma[k, c*d] = DivisorSigma[k, c] * DivisorSigma[k, d]

              Thus, if m = p1^e2 * p2^e2 ... * pn^en is the prime factorization, we have
              DivisorSigma[k, m] =
              (p1^[k(e1+1)] - 1) / (p1^k - 1) *
              (p2^[k(e2+1)] - 1) / (p2^k - 1) *
              ...
              (pn^[k(en+1)] - 1) / (pn^k - 1).
              and DivisorSigma[2k, m] =
              (p1^[2k(e1+1)] - 1) / (p1^(2k) - 1) *
              (p2^[2k(e2+1)] - 1) / (p2^(2k) - 1) *
              ...
              (pn^[2k(en+1)] - 1) / (pn^(2k) - 1).

              The ratio of these two is ten equal to
              DivisorSigma[2k, m] / DivisorSigma[k, m] =
              (p1^[k(e1+1)] + 1) / (p1^k + 1) *
              (p2^[k(e2+1)] + 1) / (p2^k + 1) *
              ...
              (pn^[k(en+1)] + 1) / (pn^k + 1).

              Obviously, if all the exponents are even, each fraction of the product
              is an integer (the numerator factors algebraically in such case). This
              proves the "if" direction of your statement.

              The smallest counterexample to the "only if" part seems to be m=20, k=1:
              DivisorSigma[20, 2] = 1+4+16+25+100+400 = 546
              DivisorSigma[20, 1] = 1+2+4+5+10+20 = 42
              546 / 42 = 13

              Peter

              PS. Now I only have to hope that Yahoogroups do not mangle this mail
              beyond comprehensibility :-)

              --
              [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
            • Kermit Rose
              Sebastián Martín Ruiz Said: 1. Divisor Function Posted by: Sebastián Martín Ruiz sebi_sebi@yahoo.com sebi_sebi Date: Sun Aug 19, 2007 8:31 pm ((PDT))
              Message 6 of 7 , Aug 24 7:21 PM
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                Sebastián Martín Ruiz Said:


                1. Divisor Function
                Posted by: "Sebastián Martín Ruiz" sebi_sebi@... sebi_sebi
                Date: Sun Aug 19, 2007 8:31 pm ((PDT))

                Hello all:

                I have obtained this result:

                DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
                if and only if
                m is a perfect square

                DivisorSigma[k,m]=sum[d^k, d divisor of m]

                Can anyone prove this result?

                Sincerely

                Sebastian Martin Ruiz



                Kermit Said:




                If m is a perfect square then

                m = p1^a1 p2^a2 p3^a3 . . . pj^aj

                where each of the a1, a2, .... aj are even.

                If d divides m then

                d = p1^b1 p2^b2 p3^b3 . . . pj^bj

                where each b1,b2,...bj are less than or equal to the corresponding
                a1,a2,...aj.

                The
                DivisorSigma[k,m]=sum[d^k, d divisor of m]

                is product of

                ( 1 + p1^k + p1^(2k) + p1^(3k) + . . . + p1^(a1*k) )
                (1 + p2^k + p2^(2k) + p2^(3k) + . . .+ p2^(a2*k) )
                . . .
                (1 + pj^k + pj^(2k) + pj^(3k) + . . . + pj^(aj * k) )


                =

                ( p1^( a1 * k) - 1) / (p1^k - 1)
                (p2^(a2 * k) - 1)/ ( p2^k - 1)
                . . .
                (pj^(aj*k) - 1) / (pj^k - 1)

                When will this product be divisible by

                ( p1^( a1 * k/2) - 1) / (p1^k - 1)
                (p2^(a2 * k/2) - 1)/ ( p2^k - 1)
                . . .
                (pj^(aj*k/2) - 1) / (pj^k - 1)

                ?

                To see the answer more readily,

                define a1 = 2 c1, a2 = 2 c2, ...aj = 2 cj

                Then we are asking when is

                [ p1^(2 c1 * k) -1 ) ( p2^(2 c2 * k) - 1 ) . . . ( 2 pj ^ ( 2 cj * k ) - 1) ]
                / [ p1^(c1*k) - 1) ( p2^(c2 * k) - 1 ) . . . ( 2 pj ^ ( cj * k ) - 1) ]


                = [ p1 ^ (c1 * k) + 1 ) ( p2 ^ (c2 * k) + 1 ) . . . (pj * ( cj * k) + 1)

                an integer?
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