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RE reciprocal consecutive primes (another error)

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  • Jose Ramón Brox
    ... From: Jose Ramón Brox That s asymptotically true cause p(n) ~ n / log(n) -- -- p(n+1) / p(n) ~ (n+1)/n * log(n) / log(n+1) = 1
    Message 1 of 1 , Nov 3, 2005
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      ----- Original Message -----
      From: "Jose Ramón Brox" <ambroxius@...>

      That's asymptotically true cause p(n) ~ n / log(n) -->
      --> p(n+1) / p(n) ~ (n+1)/n * log(n) / log(n+1) = 1 and therefore you can find a natural
      n0 so that if n > n0, then |p(n+1)/p(n) - 1| < 1/2 --> p(n+1)/p(n) < 3/2 for every n > n0.

      ----------------------------------------------------

      It's not true that p(n) ~ n / log(n) (that would be pi(n) ), but p(n) ~ n * log(n)

      The procedure remains the same because p(n+1) / p(n) ~ (n+1)*log(n+1) / (n*log(n) ) ~ 1

      Jose Brox.
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