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np ( n ) = n^n + (n+1)^(n+1) ; 2 prominent questions

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  • Walter Nissen
    Greetings , all , np ( n ) = n^n + (n+1)^(n+1) has numerous interesting aspects . Two aspects of this expression are prominent . ( 1 ) : The number of prime
    Message 1 of 2 , Oct 1, 2012
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      Greetings , all ,

      np ( n ) = n^n + (n+1)^(n+1) has numerous interesting aspects .
      Two aspects of this expression are prominent .


      ( 1 ) :
      The number of prime factors of np ( n ) .
      The first line in the table of OMEGA below is :
      1 1 1 4 2 3
      The first 3 counts in the first line are each 1 ,
      because for each of n = 1 , 2 and 3 ,
      np ( n ) = n^n + (n+1)^(n+1) is prime .
      The last count in the first line is 3 , because
      6^6 + 7^7 has 3 prime factors : 11 , 239 , and 331 .

      t

      OMEGA ( n^n + (n+1)^(n+1) )

      n mod 6
      1 2 3 4 5 0

      1 1 1 4 2 3
      2 3 2 5 2 6
      4 2 3 6 3 3
      4 6 2 7 3 5
      5 3 3 5 3 3
      3 7 4 5 4 3
      2 2 2 6 3 6
      4 5 3 8 6 6
      3 5 5 5 5 3
      3 5 2 8 4 4
      6 6 6 8 8 4
      3 4 7 7 6 6
      4 4 7 9 6 7
      5 9 5 7 4 3
      5 6 7 9 3 2
      5 4 8 9 5 8
      2 4 6 11 6 5
      4 3 8 11 4 5 ( on this 18th line : 103 <= n <= 108 )

      6 7 8 7? 6 3?
      6 5 6? 11? 7 6
      6? 4? 5 7 13 7?
      5 7 10? 9 7 9
      4 6? 5? 10? 7 3
      ___ ___ ___ ___ ___ ___

      19 22 15 32 17 23 total of first 6 lines = 6 periods , 1 <= n <= 36
      65 79 81 130 77 82 total of first 18 lines = 18 periods , 1 <= n <=
      108

      3.6 4.4 4.5 7.2 4.3 4.6 average of first 18 lines

      It's notable that
      when n mod 6 = 4 , the number of prime factors is relatively large and
      when n mod 6 = 1 , the number of prime factors is relatively small and
      when n mod 6 != 1 or 4 , the number of prime factors is moderate ,
      when n mod 3 != 1 , the number of prime factors is moderate .

      Perhaps , it's easy to understand the surplus of prime factors for
      4 mod 6 , because there are a few known algebraic factors :
      3 and ( n^2 + n + 1 ) / 3 ( twice ) .
      E.g. , 4^4 + (4+1)^(4+1) = 3 * 7 * 7 * 23 ;
      Richard Guy mentioned this surplus in a private communication .
      But , in the leftmost column ,
      why are there so few prime factors for 1 mod 6 ?


      ( 2 ) :
      It's notable that p | np ( n ) , for certain
      n mod ( p ( p - 1 ) ) .
      E.g. , for p = 3 , p(p-1) = 6 , and for n mod 6 = 4 :
      np ( 4 ) = 4^4 + (4+1)^(4+1) = 3 * 7 * 7 * 23
      np ( 10 ) = 3 * 37 * 37 * 8017 * 8969
      np ( 16 ) = 3 * 7 * 7 * 13 * 13 * 34041259347101651
      The top of the p(p-1) table is :
      (2 | np ( n ) iff n mod 2 = )
      3 | np ( n ) iff n mod 6 = 4
      5 | np ( n ) iff n mod 20 = 1 8
      7 | np ( n ) iff n mod 42 = 4 12 16 33
      11 | np ( n ) iff n mod 110 = 6 7 20 34 61 62 63 101
      13 | np ( n ) iff n mod 156 = 16 22 24 53 86 94 100 147
      E.g. , np ( 21 ) = 5 * 69454092876521107983605569601
      np ( 28 ) = 3 * 5 * 271^2 * 2360926164108571968813424783598971267
      To find factors of np ( n ) , is it possible to compute this table
      faster than using ECM or NFS ?


      Over the last decade or so , a number of people have contributed to the
      pursuit of primality in this expression , to whom I'm quite grateful ;
      on the math side , notably David Broadhurst , Mike Oakes and some of the
      usual suspects ; and on the computational side , notably
      Rich Dickerson , Lionel Debroux ( recently importantly ) , yoyo@home
      ( recently importantly ) , Andy Steward , GMP-ECM , Primo , msieve and
      ggnfs ( these lists are inadequate ) .

      Cheers ,

      Walter

      ---

      Here are a few details :
      The only known prime np are for n = 1 , 2 and 3 .
      np ( 415 ) is easy to factor , it has only 2 factors and one of
      them is 29 . ( ! )
      np ( 3754 ) has at least 19 prime factors .
      Using many years of CPU , all np have been completely factored
      for n <= 111 :
      http://upforthecount.com/math/nnp2.txt ( unpublished )
      The latest updates :
      http://factordb.com/index.php?query=n^n%2B(n%2B1)^(n%2B1)&n=1&perpage=150
      Mind-bending ; states non-trivial facts about numbers too large to
      represent in the known universe ;
      but nevertheless , intended for a general audience :
      http://upforthecount.com/math/nnnp1np1.html
      Many details :
      http://upforthecount.com/math/nnp.html ( unpublished )
      http://www.primenumbers.net/prptop/searchform.php?form=(x^x%2By^y)?&action=Search
      Many messages right here :
      http://tech.groups.yahoo.com/group/primenumbers
    • Walter Nissen
      Hi , Those of you who found the columns in the earlier post difficult to see may like the spacing in this version better :
      Message 2 of 2 , Oct 2, 2012
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        Hi ,

        Those of you who found the columns in the earlier post
        difficult to see may like the spacing in
        this version better :
        http://upforthecount.com/math/nnpomega.html
        It's also linked on the home page at :
        http://upforthecount.com/
        I'd like to know how best to avoid this problem .
        pre inside angle brackets ?

        Thanks ,

        Walter
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