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19433Re: Bounds on the (2*N)th Prime

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  • w_sindelar@juno.com
    Jun 17, 2008
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      Sindelar wrote in primenumbers message 19425:
      <… Take the expression ((4*P+3*N)/2), where P represents a prime>2 and N
      represents the number of primes from and including 2 to and including P.
      Then for certain values of P and its corresponding N, the expression will
      evaluate to a prime Q, where the number of primes from and including 2 to
      and including Q is equal to (2*N). The Q's are symmetrically located
      between (2*P+N) and 2*(P+N). The D's are symmetrically located between N
      and 2*N.
      There seems to be an inexhaustible supply of such P's. Here is a partial
      list of the first 10 of such P's. (43, 163, 373, 397, 491, 1997, 2339,
      4691, 7331, 12149), and the list of corresponding N's. (14, 38, 74, 78,
      94, 302, 346, 634, 934, 1454)...>
      David Broadhurst wrote to Sindelar:
      <<I believe that there is /no/ solution to
      4*prime(n) + 3*n = 2*prime(2*n)
      for n > 352314
      1) the asymptotic value of
      R(n) = (2*prime(2*n) - 4*prime(n))/n
      R(infinity) = 4*log(2) = 2.7725887...
      by the Prime Number Theorem.
      2) For small n, there are excursions with R(n) >= 3,
      but these soon cease, with
      R(352314) = 3
      being your last solution
      R(352316) = 3 + 1/176158
      being the last value of R(n) >= 3.
      3) Note that you may use
      to find R(n) for 2*n <= 10^12.
      For example:
      The 500,000,000,000th prime is 14,638,944,639,703
      The 1,000,000,000,000th prime is 29,996,224,275,833
      showing that
      R(5*10^11) = (2*29996224275833 - 4*14638944639703)/(5*10^11)
      = 2.873339985708
      is now significantly smaller than 3,
      yet still some way above the asymptote 2.7725887...
      PS: You may quote this argument in "primenumbers", if you wish. >>
      Thank you David, for taking the time to respond to my post. It was quite
      a surprise. I had not expected to see my paltry list of 10 N's expanded
      to 92 so quickly, and to cap it off, be presented with an argument that
      makes a case for there being a limit to the number of solutions to the
      equation (4*P+3*N)=2*(the (2*N)th prime). Put another way, this means
      that for any N>352314, the difference between the (2*N)th prime and twice
      the Nth prime can never equal (3*N)/2.
      I feel a bit uneasy about this, asymptotics can be tricky. I think the
      idea would be of interest to many in the group and I hope it will provoke
      a lively discussion. Best regards
      Bill Sindelar
      Click here for low prices on a huge selection of popcorn poppers!
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