Loading ...
Sorry, an error occurred while loading the content.

Re: Square factors of b^p-1 [SG double Wieferich sequence]

Expand Messages
  • djbroadhurst
    Prime q=2*p+1 with primes b
    Message 1 of 81 , Oct 4, 2011
    • 0 Attachment
      Prime q=2*p+1 with primes b<c<p such that q^2|b^p-1 and q^2|c^p-1

      555383, 1767407, 2103107, 7400567, 12836987, 14668163, 15404867,
      16238303, 19572647, 25003799, 26978663, 27370727, 35182919,
      36180527, 38553023, 39714083, 52503587, 53061143, 53735699,
      55072427, 63302159, 70728839, 77199743, 77401679, 86334299,
      97298759, 97375319, 103830599, 106208783, 106710287, 108711599,
      112590683, 120441239, 124581719, 126236879, 128538659, 129881603,
      133833983, 141132143, 141194387, 145553399, 151565087

      Comments: (p,q) is a Sophie Germain prime pair; (b,q) and (c,q)
      are Wieferich prime pairs; each of (b,c) is a square modulo q^2.
      The sequence is now complete up to the 42nd term, q=151565087.
      Mike Oakes set a puzzle on a more general case with primes such
      that b<p1<q, c<p2<q, b<c, q^2|b^p1-1 and q^2|c^p2-1. His sequence
      is complete only up q=27370727, containing only two new known
      primes, q=2452757 and q=22796069, with p1=p2=(q-1)/4, found in
      http://tech.groups.yahoo.com/group/primenumbers/message/23175 by
      a simple analysis of http://www.cecm.sfu.ca/~mjm/WieferichBarker

      David Broadhurst, 4 October 2011, with thanks to Mike Oakes
    • 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
      • 0 Attachment
        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
      Your message has been successfully submitted and would be delivered to recipients shortly.