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19447Bounds on the (2*N)th Prime

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  • w_sindelar@juno.com
    Jun 23, 2008
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      Thanks David for further explanation of your reasoning and Andrey
      Kulsha's graphic. I've done a lot of calculating and believe everything
      strongly suggests that the following statement may be true.
      Let P represent the Nth prime and Q represent the (2*N)th prime.
      Let K represent a rational number between the integers 1 and 2.
      Then the equation (Q-2*P)=K*N always has a least one solution. The
      maximum number of solutions possible is when K=3/2.
      For those in the group that may be interested, here is David's reason for
      believing that the equation (4*P+3*N)=2*(the (2*N)th prime), which is
      equivalent to (Q-2*P)=K*N with K=(3/2), has a limited number of
      solutions. (David calculated 92)
      > 1) First, I worked out the second term in the asymptotic
      > expansion of
      > R(n) = (2*prime(2*n) - 4*prime(n))/n
      > as here:
      > R(n) ~ 4*log(2)*(1 + 1/log(n))
      > 2) Then I proved, by explicit computation,
      > that there is no solution to R(n)=3
      > with 1500000 > n > 352314 (file attached)
      > 3) In the course of this proof, I saw, by outputting
      > [n, R(n), 4*log(2)*(1+1/log(n))]
      > at intervals of 10,000
      > that the actual value, R(n), oscillates only modestly
      > about the asymptote 4*log(2)*(1+1/log(n))
      > so I believe that I have cause to believe
      > (yet cannot prove) that n=352314 is the last solution with
      > R(n)=3.
      > David
      > PS: Feel free to share this, if you might like to...
      Thanks folks. Anyone care to comment?
      Bill Sindelar
      Lower rates for Veterans. Click for VA loan information.
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