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(k*M[n])^2+1

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  • Robert
    Some results from my sojourn into the world of primes of the form P=(k.M[n])^2+1, where M[n]=2*p[1]*p[2]...p[n], where p[] are the pythagorean primes (primes
    Message 1 of 1 , May 18, 2006
      Some results from my sojourn into the world of primes of the form
      P=(k.M[n])^2+1, where M[n]=2*p[1]*p[2]...p[n], where p[] are the
      pythagorean primes (primes which are 1mod4); and k is an integer

      for k =1, P is prime at M[0], that is P=2^2+1=5
      M[5],i.e. P= (1*2*5)^2+1 =101 is prime
      M[13], i.e. (1*2*5*13)^2+1 is prime
      M[17], M[101],M[181],M[277],M[593],M[641],M[1733] are prime.. all M
      checked to M[10009].

      I have also checked the first instance primes, in (k.M[n])^2+1 by
      increasing the value of k, for all M[n] up to and including n=4993.

      As expected, there is no formula, the pattern is a generally
      increasing value of k.

      Champions, where k is the highest value seen to date are at:
      M[n] k
      29 6
      37 8
      41 18
      73 47
      389 50
      397 97
      433 103
      541 113
      617 141
      937 154
      941 170
      1249 200
      1297 272
      1381 670
      2473 875
      3457 877
      4481 1091

      Regards

      Robert Smith
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