Re: [PrimeNumbers] Re: tuplets and prime density conjecture

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• There cannot be another: The first cluster (3,5,7,11,13,17) cannot recur because the initial triple of primes (3,5,7) = (n, n+2, n+4) must contain a multiple
Message 1 of 12 , Oct 29, 2001
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There cannot be another:

The first cluster (3,5,7,11,13,17) cannot recur because the
initial triple of primes (3,5,7) = (n, n+2, n+4) must
contain a multiple of 3. The only way to arrange 6 primes
into such a dense cluster without having such a triple is
the second cluster (5,7,11,13,17,19) = (n, n+2, n+6, n+8,
n+12, n+14) which must contain a multiple of 5. To see why,
take each term mod 5.

Mark Underwood wrote:
>
> Hi Tavai
>
> That seems intuitive to me as well, so I was surprised that for 6
> consecutive primes it *seems* that (3,5,7,11,13,17) and
> (5,7,11,13,17,19) are tied for second densest and that there are no
> others. One would think that if there are an infinite number of six
> primes at this density then another could be found fairly easily. Do
> you know of another in this case?
>
> Mark
>
> --- In primenumbers@y..., "Tavai A. Mitrytu" <jtpk@y...> wrote:
> > I have a similar running conjecture which states:
> > "Any finite arrangement of primes which occurs at
> > least twice repeats infinitely." You have to have a
> > repeat to get a minimum period.
>
>
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>
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>
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-- Dan Morenus (dmorenus@...)

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