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RE odd perfect number

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  • Jose Ramón Brox
    [I tried to send this email several times these days. I hope this is the good one] ... First time I see it. Is it a result of yourself? Do you have a proof? If
    Message 1 of 3 , Sep 8, 2006
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      [I tried to send this email several times these days. I hope this is the good one]

      >I posted it, but it didn't go.
      >Look at the formula...
      >sigma(abundant#) = 2*sigma(perfect#) + associated Mp + 1

      First time I see it. Is it a result of yourself? Do you have a proof?

      If you have it, I bet it involves even perfect numbers (because of that Mp), so why do you
      expect it to be applicable to odd perfects?

      The only related result I know says that if N is a perfect or abundant number, then k·N is
      abundant, but the converse theorem is not true, if k·N is an abundant number, then N
      hasn't to be neither abundant nor perfect (think about 945 and 945/3).

      Regards. Jose Brox
    • Peter Kosinar
      ... 1) Obviously, if A is even perfect number, it must be of the form A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is straightforward,
      Message 2 of 3 , Sep 9, 2006
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        >> I posted it, but it didn't go.
        >> Look at the formula...
        >> sigma(abundant#) = 2*sigma(perfect#) + associated Mp + 1
        >
        > First time I see it. Is it a result of yourself? Do you have a proof?

        1) Obviously, if A is even perfect number, it must be of the form
        A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is
        straightforward, see http://primes.utm.edu/notes/proofs/EvenPerfect.html.
        Then, sigma(A) = 2^n*M_n and sigma(2A) = (2^(n+1) - 1)*2^n = 2*sigma(A) +
        (2^n-1) + 1.

        So, the formula holds for abundant numbers equal to twice-an-even-perfect.

        2) There are -way- too many abundant numbers which are NOT of the form
        2*perfect number. 18, 20, ... The formula does not (and also -can- not)
        hold for them.

        > If you have it, I bet it involves even perfect numbers (because of that
        > Mp), so why do you expect it to be applicable to odd perfects?

        In fact... WHAT exactly is "associated Mp" in the case of odd perfect
        numbers?

        Peter

        --
        [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
      • Jose Ramón Brox
        ... I was aware of that. I was just asking for a proof of the claim as it was first stated, involving all perfect numbers (even the theoretical odd ones!). I m
        Message 3 of 3 , Sep 9, 2006
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          >1) Obviously, if A is even perfect number, it must be of the form
          >A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is
          >straightforward, see http://primes.utm.edu/notes/proofs/EvenPerfect.html.
          >Then, sigma(A) = 2^n*M_n and sigma(2A) = (2^(n+1) - 1)*2^n = 2*sigma(A) +
          >(2^n-1) + 1.

          I was aware of that. I was just asking for a proof of the claim as it was first stated,
          involving all perfect numbers (even the theoretical odd ones!). I'm also aware that this
          proof can't be achieved... it was just a rethorical message, a wait-a-minute-and-think
          calling for the author.

          >So, the formula holds for abundant numbers equal to twice-an-even-perfect.

          Yes, I know :-)

          >2) There are -way- too many abundant numbers which are NOT of the form
          >2*perfect number. 18, 20, ... The formula does not (and also -can- not)
          >hold for them.

          Any k·even_perfect(n) is an abundant number, in fact.

          Regards. Jose Brox
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