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• ## Re: [PrimeNumbers] Gordon Lee Puzzle...

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• I didn t asume that the primes separated by six must be consecutive. It seemed somewhat unnatural to me to exclude, for example, the pair: (1091, 1997) because
Message 1 of 3 , May 27, 2003
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I didn't asume that the primes separated by six must be consecutive.

It seemed somewhat unnatural to me to exclude, for example, the pair: (1091,
1997) because the mere fact that (1087, 1093) exists.

Josechu
josechu.com

>
> Thanks for that data Josechu. I see that like Jud you are assuming
> the primes separated by six are consecutive. Jud's data shows a
> similar ratio for B_6/B_2 as about .1813/.1043 = 1.738, and this is
> for primes between 10^4 and 10^11. This shows me that as the prime
> range gets higher, more and more primes separated by 6 are
> consecutive, that is, not part of 3 tuples. The increasing nature of
> your figures show that the 3 tuples are getting more and more rare.
> Perhaps at very, very high sample areas B_6/B_2 will approach the
> number 2 as 3 tuples become relatively infinitely rare?
>
> Mark
>
>
> --- In primenumbers@yahoogroups.com, "Josechu" <josechu8@w...> wrote:
> > I have tested the quotient B_6 / B_2 with a QuickBasic program:
> >
> > For primes between 10^3 and 2*10^4
> > B_6/B_2 = 1.6827....
> >
> > For primes between 10^3 and 4*10^4
> > B_6/B_2 = 1.7006....
> >
> > For primes between 10^3 and 6*10^4
> > B_6/B_2 = 1.70713....
> >
> > For primes between 10^3 and 8*10^4
> > B_6/B_2 = 1.7142....
> >
> > For primes between 10^3 and 10*10^4
> > B_6/B_2 = 1.713....
> >
> > For primes between 10^3 and 12*10^4
> > B_6/B_2 = 1.7131....
> >
> > For primes between 10^3 and 14*10^4
> > B_6/B_2 = 1.715694....
> >
> > For primes between 10^3 and 16*10^4
> > B_6/B_2 = 1.71569....
> >
> >
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