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Re: prime_n^(1/n) - prime_(n+1)^(1/(n+1))

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  • djbroadhurst
    ... which was an unhelpful way of defining a(k) = 2 - prime(k+1)^(1/(k+1)) At large k, the PNT gives (1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k)) David
    Message 1 of 4 , Feb 2, 2012
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      --- In primenumbers@yahoogroups.com,
      "John" <reddwarf2956@...> wrote:

      > a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1)))

      which was an unhelpful way of defining

      a(k) = 2 - prime(k+1)^(1/(k+1))

      At large k, the PNT gives

      (1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k))

      David
    • djbroadhurst
      ... but meant to write (1 - a(k)) = (log(k)/k)*(1 + O(log(log(k))/log(k)))
      Message 2 of 4 , Feb 2, 2012
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote:

        > a(k) = 2 - prime(k+1)^(1/(k+1))
        > (1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k))

        but meant to write

        (1 - a(k)) = (log(k)/k)*(1 + O(log(log(k))/log(k)))
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