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HARDY LITTLEWOOD CONJECTURE ANALISYS. FIRST PROPOSITION.

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  • ALBERTO
    Let p any prime of the form 6k+-1 and n any integer number and k the k value corresponding to each prime of the form 6k+-1 (for instance k values corresponding
    Message 1 of 3 , Jan 5, 2011
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      Let p any prime of the form 6k+-1
      and n any integer number
      and k the k value corresponding to each prime of the form 6k+-1
      (for instance k values corresponding to prime 5 is 1 because 5=(6*1)-1
      and k value for 109 is 18 because 109=6*18+1)
      Therefore the algebraic expression
      pn+-k is equivalent to 6ab+-a+-b
      In other words if an integer x is not equat to pn+-k it won't be equal to 6ab+-a+-b
      I wish to know if in your opinion this proposition is right or wrong.
      Kind regards,
      Alberto Zelaya
    • djbroadhurst
      ... That was utter nonsense, since 17 = 6*3 - 1 is prime and 31 = 17*2 - 3. Abject apologies for relying on dud code, without performing a mental sanity check.
      Message 2 of 3 , Jan 6, 2011
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote nonsense:

        > There is no prime
        > p = 6*k +/- 1 such that
        > 31 = p*n +/- k with the signs matching.

        That was utter nonsense, since
        17 = 6*3 - 1 is prime and
        31 = 17*2 - 3.

        Abject apologies for relying on dud code,
        without performing a mental sanity check.

        David
      • djbroadhurst
        ... Let s try to parse that carefully. (I failed to do so, previously, sorry.) Proposition: Let x be a positive integer for which there exists at least one
        Message 3 of 3 , Jan 6, 2011
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          --- In primenumbers@yahoogroups.com,
          "ALBERTO" <arfzelaya@...> wrote:

          > Let p any prime of the form 6k+-1
          > and n any integer number
          ...
          > if an integer x is not equal to pn+-k
          > it won't be equal to 6ab+-a+-b

          Let's try to parse that carefully.
          (I failed to do so, previously, sorry.)

          Proposition: Let x be a positive integer for which there exists
          at least one triplet (a,b,s) with integers a >= b > 0
          and a sign s = +/-1 such that x = 6*a*b + s*(a+b).
          Then there exists at least one such triplet for which
          at least one of 6*a + s and 6*b + s is prime.

          Proof: 6*x + 1 = (6*a + s)*(6*b + s) has at least one prime divisor.

          Example 1: With x = 1627604, there are 5 triplets,
          yielding only 1 prime, since 6*x + 1 = 5^10.

          Example 2: With x = 6197024, there are 63 triplets,
          yielding 7 primes, since 6*x + 1 = 23#/6.

          Question for Alberto: What has this to do with H+L ?

          David (with apologies for earlier stupidity)
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