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

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  • Dan Morenus
    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.
      >
      >
      > Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
      > The Prime Pages : http://www.primepages.org
      >
      >
      >
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      -- Dan Morenus (dmorenus@...)

      -- This parachute is not warranted to be suitable --
      -- for any purpose, including descending safely --
      -- from an airplane to the ground. --
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