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25585Re: Exact Recurrence Formula for Prime Numbers

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  • djbroadhurst
    Jul 23, 2014
      ---In primenumbers@yahoogroups.com, <yuri1777@...> wrote :


      > However, this formula cannot be used for direct
      > calculations due to rapidly growing huge precision
      > required.

      I like your use of a truncated Euler product and a truncated
      sum for zeta(4*prime(n)) to produce this impractical formula
      for prime(n+1) given all of the first n primes. Your
      choice of the multiplier 4 in the argument of zeta neatly
      exploits Betrand's observation, proven by Chebyshev.

      "Chebyshev said it, but I'll say it again;
      There's always a prime between n and 2n."

      Very neat, though utterly useless (as you acknowledge).

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