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19429Re: Bounds on the (2*N)th Prime

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  • Mark Underwood
    Jun 14, 2008
      --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
      >
      > Let P represent an odd prime>3 and N represent the number of primes from
      > and including 2 to and including P. Then in the set of positive integers
      > beginning with (2*P+N) and ending with 2*(P+N), there exists a prime Q
      > where the number of primes from and including 2 to and including Q is
      > equal to (2*N). The difference D=(Q-2*P) is always a number between
      N and
      > 2*N.
      > For example, take P=47. N=15. (2*P+N)=109. 2*(P+N)=124. The (2*N)th
      prime
      > is 113. It lies between 109 and 124. D=(Q-2*P)=19. It lies between N=15
      > and 2*N =30.
      > Here is a neat consequence of the above statement. 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).
      > The only web reference I was able to find for this, is the
      > Bertrand-Chebyshev theorem, which says that there is always at least one
      > prime between N and 2*N-2. I can see no connection. I have been
      trying to
      > prove that it is an obvious consequence of the prime number theorem, but
      > can't see my way clear. Can anyone help?
      > Thanks folks. Any comments would be appreciated.
      > Bill Sindelar


      Someone's probably privately replied already, but yes your result is a
      consequence of the prime number theorem (although it certainly doesn't
      prove that your result will always hold):

      The nth prime occurs around n*log(n), which is close to your p.

      The (2*n)th prime occurs around 2*n*log(2*n).

      But log(2*n) = log(2) + log(n), so the (2*n)th prime occurs around

      2*n*log(n) + 2*n*log(2) =~ 2*n*log(n) + 1.386*n

      But n*log(n) is close to your p, so the (2*n)th prime occurs around

      2*p + 1.386*n, which is in between 2*p + n and 2*p + 2*n as you have
      found.

      Mark

      .
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