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Re: [PrimeNumbers] Re: Number of factors

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  • Bernardo Boncompagni
    ... Interestingly enough, if p 1, sum=B_1+log(log(p)) and we get log(log(N))-log(log(p)). The chance of N being prime is 1/log(N). Invoking Mertens 3rd
    Message 1 of 7 , Oct 3, 2011
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      >
      > Provided that p << N, I guess that the average value
      > of omega(n/p) for the numbers n up to N with least
      > prime divisor p is
      >
      > log(log(N)) + B_1 - sum(prime q < p, 1/q) + o(1)
      >
      > with B_1 =
      > 0.261497212847642783755426838608695859051566648261199206192...
      >
      Interestingly enough, if p>>1, sum=B_1+log(log(p)) and we get
      log(log(N))-log(log(p)).

      The chance of N being prime is 1/log(N). Invoking Mertens' 3rd theorem, the
      chance of a number sieved up to p being prime should be
      log(p)/(log(N)*exp(-gamma)), where gamma is the Euler-Mascheroni constant.
      If the log(N) in the number of divisors is distantly related to the chance
      of N being prime, one may squeeze this result into our formula obtaining
      log(log(N)*exp(-gamma)/log(p))=log(log(N))-log(log(p))-gamma. Not sure if
      this last passage makes sense but at this point I guess
      log(log(N))-log(log(p)) is a very reasonable answer to the problem. By the
      way, N>p^2 means log(log(N))-log(log(p))>log(2) which looks reasonable
      enough.

      It's even very elegant if one writes it as log(lg(N)), where lg is the
      logarithm to base p...


      [Non-text portions of this message have been removed]
    • djbroadhurst
      ... For log(p)
      Message 2 of 7 , Oct 3, 2011
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        --- In primenumbers@yahoogroups.com,
        Bernardo Boncompagni <RedGolpe@...> wrote:

        > > log(log(N)) + B_1 - sum(prime q < p, 1/q) + o(1)
        >
        > Interestingly enough, if p>>1, sum=B_1+log(log(p))
        > and we get log(log(N))-log(log(p)).

        For log(p) << log(N), the average of omega(n/p),
        with n running from p to N and p the least
        prime divisor of n, is (I claim) asymptotic to the
        sum of 1/q, with prime q running from q = p up to
        the greatest prime q <= N/p.

        All I did was to use Mertens at the upper limit.
        For log(2) << log(p) << log(N), one may also use
        Mertens at the lower limit to get log(log(N)/log(p)),
        as Bernardo has observed. However one should not push
        this up to p = O(sqrt(N)), since there we may run into
        a problem that worried Chebyshev:
        http://tech.groups.yahoo.com/group/primenumbers/message/21565

        David (per proxy Pafnuty Lvovich)
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