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Brun's constant in a different context

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  • anastasis_1999
    hi so I ve been trying for awhile to write an asymptotic formula for sum p_n
    Message 1 of 2 , Apr 2, 2014
      hi
      so I've been trying for awhile to write an asymptotic formula for
      sum p_n<=x 1/(p_n - p(n-1))
      (the sum of reciprocals of prime gaps)
      where p_n means nth prime and n>1.

      my attempt is this
      x/ln(x)^2*(A*ln(ln(X))+(B*C_twin))+err.term, C_twin=0.66016...
      where A=approx 0.37787 and B is Brun's constant.
      The value of A is ln(2*C_twin)+1/10, and the error term is very small even for large x. Can someone verify this please? I can send you a spreadsheet if you like, I've tested this up to 500,000..
      thanks from Guy
    • djbroadhurst
      ... At large x, the probability of primality is asymptotic to 1/log(x) and hence the average gap between primes is log(x). The integral of 1/log(x)^2 diverges
      Message 2 of 2 , Apr 3, 2014
        "Punkish301" wrote:

        > asymptotic formula for
        > sum p_n<=x 1/(p_n - p(n-1))

        At large x, the probability of primality is asymptotic to
        1/log(x) and hence the average gap between primes is log(x).
        The integral of 1/log(x)^2 diverges as x/log(x)^2.

        However, your sum favours small gaps. So let us consider the
        role of the twin primes. They have a conjectural density of
        order 1/log(x)^2. Hence their contribution to the sum is, by
        itself, of order x/log(x^2). So one might expect that the
        full sum diverges faster than x/log(x)^2.

        > I've tested this up to 500,000.

        We may easily go further.In a few minutes,
        I obtained results up to x = 10^10.

        For prime p > 2, let g(p) = p - q be the gap between p and
        the previous prime, q. Then consider this sum over prime p:
        S(x) = sum(2 < p < x, 1/g(p))

        Here is a table of [x, S(x)*log(x)^2/x]:

        [1000, 1.932409314]
        [10000, 2.104903983]
        [100000, 2.175330374]
        [1000000, 2.252107255]
        [10000000, 2.335029340]
        [100000000, 2.402612252]
        [1000000000, 2.468588116]
        [10000000000, 2.530925912]

        So it is likely that S(x) increases faster than x/log(x)^2.

        David
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