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RE distance between prime tuples

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  • Jose Ramón Brox
    Mark said: Prime triples of the form p,p+2,p+6 are separated from adjacent similar triples of the same form by multiples of 6. Multiples such as 1*6, 4*6,5*6,
    Message 1 of 1 , May 30, 2005
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      Mark said:

      Prime triples of the form p,p+2,p+6 are separated from adjacent
      similar triples of the same form by multiples of 6. Multiples such
      as 1*6, 4*6,5*6, 6*6, 8*6, and so on. However as far as I can tell,
      they are never separated from each other by 2*6, 3*6, 7*6, and
      perhaps some other multiples of 6 which I haven't looked for. Offhand
      I don't know why this would be. Anybody want to hazard a guess? Same
      thing goes for triples of the form p,p+4,p+6 I think.

      -------------------------------------------

      Take a triplet (p,p+2,p+6):

      If p+1 is a multiple of 5, then we have two cases: p==4 (mod 10) or p== 9 (mod 10).
      In the first case, since gcd(4,10)=2, p==0 (mod 2) and p is composite. In the second case,
      p=k*10+9 and p+6 = k*10+9+6 = (k+1)*10+5 ==0 (mod 5). That's why you'll never see a
      triplet starting with a prime with nine as last digit (see
      http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A022004 ).

      If p+3 is a multiple of 5, then so it is p+3+15 = p+6+2*6

      If p+4 is a multiple of 5, then so it is p+4+10 = p+2+2*6

      (p+5 can't be a multiple of 5, or p will be another one)

      As we have seen, the number 5 stops all our intents to get (p+2*6,p+2+2*6,p+6+2*6) as a
      prime triplet. We only have a special case: if we allow p==0 (mod 5), then we get
      (5,7,11) --> (17,19,23) that are two prime triplets separated by 2*6! ;-)

      I think the rest of your examples can be resolved the same way, but I have an exam
      tomorrow... I'm sure you can follow the idea to see if it remains explaining the
      phenomenon (with 5, 7, combinations... whatever).

      Regards. Jose Brox.
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