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Re: Square factors of b^p-1

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  • mikeoakes2
    ... New experimentally-determined data point supports the picture:- 10^7.5 1944 2.489 Mike
    Message 1 of 81 , Oct 3, 2011
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      --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@> wrote:
      > >
      > > Puzzle 2: Find a triplet of primes (b,p,q) with b < p < q,
      > > b^p = 1 mod q^2, and q^2 having precisely 21 decimal digits.
      >
      > I reckon there should be about 208952 solutions to this puzzle.
      > Rationale:-
      >
      > max_q solutions ratio
      > 10^1 0 -
      > 10^1.5 1 -
      > 10^2 1 1.0
      > 10^3 1 1.0
      > 10^3.5 1 1.0
      > 10^4 3 3.0
      > 10^4.5 6 2.0
      > 10^5 22 3.667
      > 10^5.5 58 2.636
      > 10^6 125 2.155
      > 10^6.5 322 2.576
      > 10^7 781 2.435
      > Taking the ratio to be about 2.4 then gives these estimates:-
      > 10^7.5 1874 2.4
      >...

      New experimentally-determined data point supports the picture:-
      10^7.5 1944 2.489

      Mike
    • Walter Nissen
      Hi , all , 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
      Message 81 of 81 , Nov 29, 2011
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        Hi , all ,

        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 ( 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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