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Divisor Function

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  • 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 1 of 7 , Aug 24, 2007
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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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