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prime_n^(1/n) – prime_(n+1)^(1/(n+1))

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  • John W. Nicholson
    a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) ,  {n,1,k} means n=1 to k. For k 1.   What is the largest
    Message 1 of 4 , Feb 2, 2012
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      a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) ,  {n,1,k} means n=1 to k.
      For k <= 1,000,000, observe a < 1, and a-->1.

       
      What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?
       
      John W. Nicholson

      [Non-text portions of this message have been removed]
    • John
      I do not know why the extra characters were place in, but here it is again. a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to
      Message 2 of 4 , Feb 2, 2012
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        I do not know why the extra characters were place in, but here it is again.

        a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to k.
        For k <= 1,000,000, observe a < 1, and a-->1.

        What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

        John W. Nicholson


        --- In primenumbers@yahoogroups.com, "John W. Nicholson" <reddwarf2956@...> wrote:
        >
        > a := sum ({n,1,k}, prime_n^(1/n) â€" prime_(n+1)^(1/(n+1))) ,  {n,1,k} means n=1 to k.
        > For k <= 1,000,000, observe a < 1, and a-->1.
        >
        >  
        > What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?
        >  
        > John W. Nicholson
        >
        > [Non-text portions of this message have been removed]
        >
      • 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 3 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 4 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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