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twin-prime heuristic proof

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  • Bill Bouris
    Hello, Group. I m optimistic that no one will find a counter-example. This method is similar to Euclid s proof for infinite-ly many primes. Let k = 2, 3, 4, 5,
    Message 1 of 3 , Jul 8 12:24 PM
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      Hello, Group.

      I'm optimistic that no one will find a
      counter-example.

      This method is similar to Euclid's proof for
      infinite-ly many primes.

      Let k = 2, 3, 4, 5, etc. and then...
      compute z = [(p(sub k) primorial) / 2] + 2^k.

      Take a look at the results...

      z -2 = 7 - 2 = 5 (twin-prime number)
      z = 3 + 2^2 = 7 (also twin-prime)
      z +2 = 7 + 2 = 9 (analysis not necessary)

      Clearly, the twin-prime pair is 5,7 and 5 > 3.

      Continuing...

      z -2 = 21 = 3 * 7 (again the pair is 5,7 and 7 > 5)
      z = 3*5 + 2^3 = 23 (prime but not important)
      z +2 = 23 + 2 = 25 (factorization not needed)

      Once again...

      z -2 = 119 (analysis not necessary)
      z = 3*5*7 + 2^4 = 121 = 11 * 11; (the twin-prime
      pair is 11,13 and 11 > 7)
      z +2 = 123 (ignore this number too)

      Either the pair z-2,z or z,z+2 will be a twin-prime
      pair or one of the three will contain a factor from a
      twin-prime pair that is > p(sub k).

      I checked the results up to p(sub 10) = 29.

      Also, I found a variation of the formula for primes
      that are just four units apart.

      Has this relationship already been found? ... or does
      someone find it interesting???

      Bill




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    • Peter Kosinar
      Hello Bill, ... Unfortunately, your conjecture seems to fail for k=29 prime(29) = 109 primorial(109)=279734996817854936178276161872067809674997230
      Message 2 of 3 , Jul 8 3:45 PM
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        Hello Bill,

        > I'm optimistic that no one will find a counter-example.

        Unfortunately, your conjecture seems to fail for k=29

        prime(29) = 109
        primorial(109)=279734996817854936178276161872067809674997230
        corresponding z=139867498408927468089138080936033905374369527

        1) z-2 =
        3*5^2*29*43*1495509205120849698894820432355347825441

        The small factors do not qualify as they're smaller than 109 and the big
        one (t=1495509205120849698894820432355347825441) does not belong to a twin
        prime pair (t-2 is divisible by 29, t+2 by 3).

        2) z =
        373*16159597*3217431548819*7212206181309313175293 = P1*P2*P3*P4

        No twin primes here, as
        P1+2, P2+2, P3-2, P4+2 are divisible by 3,
        P2-2 is divisible by 5,
        P1-2 and P3+2 are divisible by 7,
        and P4-2 by 29.

        3) z+2 =
        17*307*1559*10259*393083*4262791622718255614129772517=P1*P2*P3*P4*P5*P6

        P1 does not qualify, the others aren't members of twin prime pairs, as

        P2+2, P3-2, P4-2, P5-2, P6+2 are divisible by 3,
        P2-2, P5+2, P6-2 are divisible by 5,
        P3+2 is divisible by 7,
        P4+2 is divisible by 31.

        Did I do anything wrong?

        > Has this relationship already been found? ... or does
        > someone find it interesting???

        It is undoubtedly nice but I guess it's just a phenomenon
        for small numbers (as the twin primes are quite common in that region).

        Peter

        --
        [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
      • Jens Kruse Andersen
        ... No. PARI/GP says the first counter examples are for k = 29, 33, 38, 39. ... That s an awfully short check, especially for a form allowing so many twin
        Message 3 of 3 , Jul 8 4:34 PM
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          Peter Kosinar wrote:
          > Unfortunately, your conjecture seems to fail for k=29
          ...
          > Did I do anything wrong?

          No. PARI/GP says the first counter examples are for k = 29, 33, 38, 39.

          Bill Bouris wrote:
          > I checked the results up to p(sub 10) = 29.

          That's an awfully short check, especially for a form allowing
          so many twin chances.
          There is no justification for the words "heuristic" or "proof".
          It was just a poor guess.

          --
          Jens Kruse Andersen
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