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Re: [PrimeNumbers] RE odd perfect number

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  • 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 1 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 2 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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