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Re: bigger prime gaps and higher prime powers [Exercise]

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  • John
    ... The prime sequence has no memory If this were true, then we would see the Ramanujan primes ~ 3n instead of ~ 2n. http://arxiv.org/abs/1105.2249 The
    Message 1 of 37 , Oct 12, 2011
      --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:

      > Ok, let's try and spell it out.
      >
      > Start with an arbitrary (and large, so PNT applies) integer N.
      >
      > On either side of N, an integer has a chance 1/log(N) of being prime;
      > so if one travels (in either direction), you have to go on average log(N) before finding a prime.
      >
      > p1 < N <=p < p2.
      >
      > From N, travel downwards: it takes log(N) to find p1, so av. (N-p1)=log(N).
      > From N, travel upwards; it takes log(N) to find p, so av. (p-N)=log(N).
      > The prime sequence has no "memory", so, continuing upwards, from p to p2 takes another log(N), so av. (p2-p)=log(N).
      > Adding all 3 averages gives C=3.
      >
      > Mike
      >

      "The prime sequence has no "memory""

      If this were true, then we would see the Ramanujan primes ~ 3n instead of ~ 2n. http://arxiv.org/abs/1105.2249

      The "memory" is in a way a type of "feedback" by the prior primes to the future gap, and is related to the number of Ramanujan primes, n. See the corollary at http://en.wikipedia.org/wiki/Ramanujan_prime and think about how these log N's add up between x/2 and x. Note also Generalized Ramanujan primes and how they relate. Also note, how sharp the range is between [p_k,2*p_(i-n)]. This is where p_i is found.
    • John
      ... The prime sequence has no memory If this were true, then we would see the Ramanujan primes ~ 3n instead of ~ 2n. http://arxiv.org/abs/1105.2249 The
      Message 37 of 37 , Oct 12, 2011
        --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:

        > Ok, let's try and spell it out.
        >
        > Start with an arbitrary (and large, so PNT applies) integer N.
        >
        > On either side of N, an integer has a chance 1/log(N) of being prime;
        > so if one travels (in either direction), you have to go on average log(N) before finding a prime.
        >
        > p1 < N <=p < p2.
        >
        > From N, travel downwards: it takes log(N) to find p1, so av. (N-p1)=log(N).
        > From N, travel upwards; it takes log(N) to find p, so av. (p-N)=log(N).
        > The prime sequence has no "memory", so, continuing upwards, from p to p2 takes another log(N), so av. (p2-p)=log(N).
        > Adding all 3 averages gives C=3.
        >
        > Mike
        >

        "The prime sequence has no "memory""

        If this were true, then we would see the Ramanujan primes ~ 3n instead of ~ 2n. http://arxiv.org/abs/1105.2249

        The "memory" is in a way a type of "feedback" by the prior primes to the future gap, and is related to the number of Ramanujan primes, n. See the corollary at http://en.wikipedia.org/wiki/Ramanujan_prime and think about how these log N's add up between x/2 and x. Note also Generalized Ramanujan primes and how they relate. Also note, how sharp the range is between [p_k,2*p_(i-n)]. This is where p_i is found.
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