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Small point

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  • Milton Brown
    OK, so exp(-g/(ln x + ln(ln x))) produces 1-exp(-0.000433 g) { not 436 } I guess that this just proves, that I am using your equation correctly. Milton ...
    Message 1 of 1 , Jul 26, 2001
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      OK, so

      exp(-g/(ln x + ln(ln x)))

      produces

      1-exp(-0.000433 g) { not 436 }

      I guess that this just proves, that I am using your
      equation correctly.

      Milton


      ----- Original Message -----
      From: "Andrey Kulsha" <Andrey_601@...>
      To: <PrimeNumbers@...>
      Sent: Thursday, July 26, 2001 3:18 AM
      Subject: [PrimeNumbers] Gaps distribution: Conjecture


      > Hello!
      >
      > Milton Brown wrote:
      >
      >
      > > I have made an empirical study of prime gaps around 10^1000+k,
      > > small k, where your log x ~ 2303, and I got that the probability of
      > > a Gap of less than size g is
      > >
      > > Prob(x < g) = 1 - exp(-0.00476635 g)
      > >
      > > which says that 50% of the Gaps here occur at g <= 1454.
      > >
      > > Is it possible for you to characterize your equation this
      > > way for comparison?
      >
      > As I see, you compute the probability among gaps, but I compute f(g, p)
      among
      > numbers, i.e. there are about f(g, p)*eps gaps of a size g between x and
      x+eps,
      > where g<<eps<<x.
      >
      > log x ~ 2303, log log x ~ 8, C(g) ~ 0.7, so
      >
      > 2303*sum(exp(0.7 - (2k+1) / (2303 - 8) ) / 2303^2, k, 0, g/2-1) ~
      > ~ 1 - exp( -g/2295) ~ 1 - exp( -0.000436 g).
      >
      > so about 50% of the gaps are less than 1591.
      >
      > Note that your result is quite similar (if added a zero that you've
      missed, as I
      > guess).
      >
      > General result: the probability of a given gap g around the number x being
      > greater than G is about
      >
      > exp( -g / (log x + log(log x)) ).
      >
      > Best wishes,
      >
      > Andrey
      >
      >
      >
      >
      >
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