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RE odd perfect form

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  • Jose Ramón Brox
    ... From: leavemsg1 ... A = 3^(2n-2), B = (3^(2n-1)-1)/2 N = A·B Sigma is a multiplicative function, and B = = 1 (mod 3), which
    Message 1 of 3 , Sep 1, 2006
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      ----- Original Message -----
      From: "leavemsg1" <leavemsg1@...>


      >Sorry,... the formula should have been posted as...
      >(3^(2n-2))*(3^(2n-1)-1)/2.

      A = 3^(2n-2), B = (3^(2n-1)-1)/2
      N = A·B

      Sigma is a multiplicative function, and B = = 1 (mod 3), which implies CGD(A,B) = 1, so

      Sigma(N) = Sigma(A·B) = Sigma(A)·Sigma(B)

      Sigma(A) = 1+3+3^2 + ... + 3^(2n-2) = (3^(2n-1)-1)/2 = B

      and as we want Sigma(N) = 2N = 2AB, it follows that we need Sigma(B) = 2A

      Sigma((3^(2n-1)-1)/2) = 2·3^(2n-2), can this be?

      I thought I had a simple proof for the negative, but it was flawed. Nevertheless, let's
      try other ways. I don't have the time to prove anyone, but I will state them:

      a) I think that the only exponents that make Sigma(B) = = 2 (mod 4) are n = 2, 4, 7, 36
      and 52. This could be mistaken easily.
      b) I also believe that Sigma(B) < 2A for every n.
      c) Probably there is yet a simpler way, but I don't see it right now.

      Jose Brox
    • leavemsg1
      ... implies CGD(A,B) = 1, so ... (B) = 2A ... Forget my attempt to formalize a formula for an odd-perfect number be- cause by your very sigma definitions...
      Message 2 of 3 , Sep 2, 2006
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        --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@...>
        wrote:
        >
        >
        > ----- Original Message -----
        > From: "leavemsg1" <leavemsg1@...>
        >
        >
        > >Sorry,... the formula should have been posted as...
        > >(3^(2n-2))*(3^(2n-1)-1)/2.
        >
        > A = 3^(2n-2), B = (3^(2n-1)-1)/2
        > N = A·B
        >
        > Sigma is a multiplicative function, and B = = 1 (mod 3), which
        implies CGD(A,B) = 1, so
        >
        > Sigma(N) = Sigma(A·B) = Sigma(A)·Sigma(B)
        >
        > Sigma(A) = 1+3+3^2 + ... + 3^(2n-2) = (3^(2n-1)-1)/2 = B
        >
        > and as we want Sigma(N) = 2N = 2AB, it follows that we need Sigma
        (B) = 2A
        >
        > Sigma((3^(2n-1)-1)/2) = 2·3^(2n-2), can this be?
        >
        Forget my attempt to formalize a formula for an odd-perfect number be-
        cause by your very sigma definitions... the first perfect number of
        any even or odd number must occur at half the number of the first
        abundant number found in the list i.e. 6 is the first perfect number
        which is half of and less than 12(the first even abundant number).
        Hence, by definition... the first odd perfect number would have to
        occur before 945/2 which isn't a whole number.

        Hence, i believe an odd-perfect number defined by the sigma function
        couldn't exist.

        Bill


        > I thought I had a simple proof for the negative, but it was flawed.
        Nevertheless, let's
        > try other ways. I don't have the time to prove anyone, but I will
        state them:
        >
        > a) I think that the only exponents that make Sigma(B) = = 2 (mod 4)
        are n = 2, 4, 7, 36
        > and 52. This could be mistaken easily.
        > b) I also believe that Sigma(B) < 2A for every n.
        > c) Probably there is yet a simpler way, but I don't see it right
        now.
        >
        > Jose Brox
        >
      • Jose Ramón Brox
        ... From: leavemsg1 ... Why is that so? If you have a sound argument for that idea, i.e., that the first perfect number must occur at
        Message 3 of 3 , Sep 2, 2006
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          ----- Original Message -----
          From: "leavemsg1" <leavemsg1@...>

          >Forget my attempt to formalize a formula for an odd-perfect number be-
          >cause by your very sigma definitions... the first perfect number of
          >any even or odd number must occur at half the number of the first
          >abundant number found in the list i.e. 6 is the first perfect number
          >which is half of and less than 12(the first even abundant number).
          >Hence, by definition... the first odd perfect number would have to
          >occur before 945/2 which isn't a whole number.

          Why is that so? If you have a sound argument for that idea, i.e., that the first perfect
          number must occur at half of the first abundant number... then you could gain fame and
          fortune instantly! The Odd Perfect's Conjecture is considered to be a hard one.

          Regards. Jose Brox
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