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n^2+1 prime k-tuplets

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  • Robin Garcia
    Mike,you are right. I had very few time and miscalculated. g(10)=30 because we always can find the g(12)=38 ---------- 66,.....,104 (nod 170) incuding the
    Message 1 of 1 , Nov 2, 2004
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      Mike,you are right.
      I had very few time and miscalculated.
      g(10)=30
      because we always can find the g(12)=38 ---------->66,.....,104 (nod 170)
      incuding the 0¨s,of course. And they fit with 106,......,144 (mod 130)
      or 116,......,154 (mod 130)

      So g(10)=30
      g(11)=34
      g(12)=38
      g(13)=42
      g(14)=46
      g(15)=50
      g(16)=54

      Sometimes,like Fermat,I make silly errors (almost all the Fermat primes are composite............)
      Hope this is not again the case.

      There is a typo in my previous post :must be--->

      29 divides n^2+1 for n=46,104,186,244 (mod 290)
      37 ---------------------------n=6,154,364 (mod 370)


      "If the world was only represented in black and white ,it would seem grey
      for us.So we´ll build a grey world" (Bill Gates,again......)

      Happy binary day.



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