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Re:k difference primes theorems

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  • cino hilliard
    hi, Consider 1. If p is prime and p+k is prime then p+k divides p^(p+k) + k. 2. the converse If p is prime and p+k is composite and p+k divides p^(p+k) +k,
    Message 1 of 1 , Jan 9, 2005
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      hi,

      Consider

      1. If p is prime and p+k is prime then p+k divides p^(p+k) + k.
      2. the converse

      If p is prime and p+k is composite and p+k divides p^(p+k) +k, then p+k is a
      pseudokprime.

      For k=2 we get the case for twin primes. There are 33 pseudotwinprimes for
      1517 twin
      primes less than 130000. In other words of the 12159 primes p less than
      130000 the divide
      by p+2 test failed 33 times to detect if p and p+2 were or were not twin
      primes.

      for k=4 or quad primes, there are 11 pseudoquadprimes for 344 quadprimes
      less than 20000.

      Thus it appears, at least statistically, that it is highly probable that
      p+k is prime if p+k divides
      p^(p+k) + k.

      It is interesting to note that Charmichael numbers pop up in these
      calculations.

      Maybe someone can prove 1. and elaborate. I have included a pari script
      below.


      ktokpk(n,n2,k) =
      {
      local(x,y,x2,c);
      c=0;
      forprime(x=n,n2,
      x2=x+k;
      y=x^x2+k;
      if(y%x2==0&!isprime(x2),c++;
      print1(x2",");
      );
      );
      print();
      print(c","pikprimes(n2,k))
      }


      pikprimes(n,k) = \\The number of k difference prime pairs less than n.
      {
      local(c,x);
      c=0;
      forprime(x=3,n,
      if(isprime(x+k),c++)
      );
      return(c)
      }

      Cino
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