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Re: [PrimeNumbers] An equivalence for prime numbers

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  • Phil Carmody
    ... I m sure you ve obtained these about a dozen times in the last few years, as it s just a trivial obfuscation of Sum[z=a..b] { 1 if z|p; 0 otherwise } = #
    Message 1 of 2 , Aug 31, 2010
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      --- On Tue, 8/31/10, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      > 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.

      I'm sure you've obtained these about a dozen times in the last few years, as it's just a trivial obfuscation of

      Sum[z=a..b] { 1 if z|p; 0 otherwise } = # divisors of z in [a..b]

      z*Floor((p+z)/z) is clearly just (p+z)-((p+z)%z)

      So Floor(z*Floor((p+z)/z)/(p+z)) is clearly only 1 when (p+z)%z==0, i.e. when z|p, and otherwise 0.

      Your use of (p+z)%z rather than p%z is purely obfuscation in order to make it appear that you've come up with a new expression, but all you've done is make something no more useful, but even uglier than previous ones.

      Phil
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