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Re: [PrimeNumbers] (im)perfect numbers

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  • Ignacio Larrosa Cañestro
    Sunday, December 05, 2004 11:19 PM [GMT+1=CET], ... The sum of the first n odd cubes is S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1) If n = 2^m,
    Message 1 of 3 , Dec 6, 2004
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      Sunday, December 05, 2004 11:19 PM [GMT+1=CET],
      Michael Gian <work.gian@...> escribió:

      > As I have read, for prime numbers, p, 2^p-1 are called Prime-
      > Exponent Mersenne numbers.
      > These are sub-group of 2^k-1, k positive integers, which are called
      > (just) Mersenne numbers.
      > When 2^p-1 is prime, it is called a Mersenne prime.
      >
      > (2^(p-1)*(2^p-1)is a Perfect Number when 2^p-1 is a Mersenne prime.
      >
      > Does anyone know of a accepted term for a number of the form (2^(p-1)
      > *(2^p-1) when 2^p-1 is not prime?
      >
      > The reason I ask is that all numbers of this form, including Perfect
      > Numbers, are the sum of consecutive odd cubes. I am searching for a
      > more concise way to talk about them.
      >
      > Michael

      The sum of the first n odd cubes is

      S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1)

      If n = 2^m, then

      S(2^m) = 2^(2m)(2^(2m+1) - 1) = 2^(p-1)(2^p - 1) = M(p)

      Then Mersenne number M(p) = 2^(p-1)(2^p - 1), with odd p, but no neccesary
      prime, is the sum of the first 2^((p-1)/2) odd cubes.

      Saludos,

      Ignacio Larrosa Cañestro
      A Coruña (España)
      ilarrosa@...
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