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Re: Factors of np ( n ) = n^n + (n+1)^(n+1) ; was Square factors of n^n+(n+1)^(n+1)]

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  • Walter Nissen
    Hi , all , Correcting a typo in my message of a few minutes ago : ( n^2 + n + 1 ) / 3 Not even draft yet , let alone published :
    Message 1 of 1 , Nov 29, 2011
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      Hi , all ,

      Correcting a typo in my message of a few minutes ago :
      ( n^2 + n + 1 ) / 3

      Not even draft yet , let alone published :
      http://upforthecount.com/math/nnp.html
      still needs a lot of work .
      But , I have updated the top of the main table :
      http://upforthecount.com/math/nnp2.txt
      with complete factorizations now extending through
      np ( 90 ) .

      When is np ( n ) prime ?
      So far , only for 1 , 2 and 3 .
      When n mod 6 = 4 , np ( n ) has at least 3 algebraic
      factors , 3 and ( n^2 + n + 1 ) / 3 ( twice ) .
      But what about the residue ( primitive part ? ) ?
      When is this prime ?
      n t n^n + (n+1)^(n+1)
      4 4 3 * 7 * 7 * 23
      16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651
      Any others ?

      Cheers ,

      Walter
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