## Small point

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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 ...
Message 1 of 1 , Jul 26 8:14 AM
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@...>
Sent: Thursday, July 26, 2001 3:18 AM

> 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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