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21781An equivalence for prime numbers

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  • Sebastian Martin Ruiz
    Aug 31, 2010
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      Hello all:

      I have obtained an equivalence for prime numbres:

      p is prime

      if and only if 

      Sum {for z=1 to p^(1/2)} Floor[(z*Floor[(p+z)/z]/(p+z)] = 1

      and better:

      Sum {for z=1 to p} Floor[(z*Floor[(p+z)/z]/(p+z)] =d(p) the number of divisors
      of p.


      Sebastián Martin Ruiz

      [Non-text portions of this message have been removed]
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