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Re: [PrimeNumbers] An equation for primes

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  • Ignacio Larrosa Cañestro
    ... And it is p + g ... [h] = [p + g + O(1/p)] = p + g = q Best regards, Ignacio Larrosa Cañestro A Coruña (España)
    Message 1 of 3 , Aug 29, 2010
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      Ignacio Larrosa Cañestro wrote:
      > Sebastian Martin Ruiz wrote:
      >> Hello all:
      >>
      >> I have obtained the following equation for primes:
      >>
      >> Let p a prime number p>=127. Let g=q-p where q=the next prime to p.
      >>
      >> We have:
      >>
      >> Floor((p^(g/2+1))/((p-2)^(g/2)))=g+p=q
      >>
      >> ( floor(x) = ⌊x⌋ is the largest integer not greater than x. Example:
      >> Floor (3.459)=3)
      >>
      >
      > Dear Sebastian,
      >
      >
      > Let
      >
      > h = (p^(g/2+1))/((p-2)^(g/2)) = p(p/(p-2))^(g/2)
      >
      > = p(1/(1 - 2/p)^(g/2) ~= p(1 + 2/p + 4/p^2 + ...)^g/2
      >
      > ~= p(1 + g/p + O(1/p^2) = p + g + O(1/p)

      And it is > p + g

      >
      > Then, from certain p, h = p + g = q

      [h] = [p + g + O(1/p)] = p + g = q


      Best regards,

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