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Re: Lucas mersenne ?

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  • djbroadhurst
    Yes Shane, yours was indeed a nice observation. The classic Lucas-Lehmer test for Mersennes uses the Lucas sequence with p=2, q=-2, d=p^2-4*q=12 (Ribenboim
    Message 1 of 12 , Jan 2, 2002
      Yes Shane, yours was indeed a nice observation.

      The classic Lucas-Lehmer test for Mersennes
      uses the Lucas sequence with
      p=2, q=-2, d=p^2-4*q=12 (Ribenboim p.92).

      You can probably use the same batch of theorems
      from Ribenboim to prove the necessity and sufficiency
      of your more complicated two-case test with
      p=1, q=-1, d=p^2-q=5 (i.e. the rabbit case).

      Happy new year

      David
    • mikeoakes2@aol.com
      In a message dated 02/01/2002 08:31:16 GMT Standard Time, ... What s going on here is a special case of the following. Let q be any prime = 3 mod 4. (This
      Message 2 of 12 , Jan 5, 2002
        In a message dated 02/01/2002 08:31:16 GMT Standard Time,
        d.broadhurst@... writes:

        > > L(2^(p-1))mod 2^p -1 = 0 or, [CASE1]
        > > L(2^(p-1) -1) mod 2^p -1 = 0 [CASE2]
        >
        > p CASE
        > [ 2, CASE1]
        > [ 3, CASE1]
        > [ 5, CASE2]
        > [ 7, CASE1]
        > [13, CASE2]
        > [17, CASE2]
        > [19, CASE1]
        >
        > interesting....
        >

        What's going on here is a special case of the following.

        Let q be any prime = 3 mod 4. (This includes any of the form 2^p-1.)
        Then q divides L(m), where
        m = (q-1)/2 if q = +-1 mod 5,
        and m = (q+1)/2 if q = +-2 mod 5.
        [See e.g. Ribenboim's "New Book of Prime Number Records", Chap 2 Sec IV for a
        comprehensive treatment of all this.]

        For the Mersennes, this gives:-
        p q=2^p-1 q mod 5 m=(q+-1)/2 CASE
        2 3 -2 2 CASE1
        3 7 2 4 CASE1
        5 31 1 15 CASE2
        7 127 2 64 CASE1
        11 2047=23*89 [not prime] --
        13 8191 1 4095 CASE2
        17 131071 1 65535 CASE2
        19 524287 2 262144 CASE1

        Numbers q = 3 mod 4 which divide L(m) are not, however, NECESSARILY prime,
        only "Lucas pseudoprimes"; so we don't have a new primality test here,
        unfortunately.

        Mike Oakes
      • djbroadhurst
        Thanks for the clarification, Mike. But Shane s question was not answered by either of us: is there a Lucas mersenne pseudoprime, M(p)=2^p-1 with prime p and
        Message 3 of 12 , Jan 5, 2002
          Thanks for the clarification, Mike.
          But Shane's question was not answered by either of us:
          is there a Lucas mersenne pseudoprime,
          M(p)=2^p-1 with prime p and composite M(p)
          and one of the two tests satisfied?
          It's a nice question!
          David
        • ttpi314159
          ... So, then I most ask the most obvious question ? Using the Fibonacci sequence: Case 1: F(2^p) mod 2^p -1 = 0 Case 2: F(2^p -2) mod 2^p -1 = 0 F(4)=3 C1
          Message 4 of 12 , Jan 5, 2002
            --- In primenumbers@y..., "djbroadhurst" <d.broadhurst@o...> wrote:
            > Thanks for the clarification, Mike.
            > But Shane's question was not answered by either of us:
            > is there a Lucas mersenne pseudoprime,
            > M(p)=2^p-1 with prime p and composite M(p)
            > and one of the two tests satisfied?
            > It's a nice question!
            > David


            So, then I most ask the most obvious question ?
            Using the Fibonacci sequence:

            Case 1: F(2^p) mod 2^p -1 = 0
            Case 2: F(2^p -2) mod 2^p -1 = 0

            F(4)=3 C1
            F(8)=7 C1
            F(30)=31 C2
            F(128)=127 C1

            F(8190,8192)=8191 ?
            F( ... ?


            The cases so far between Lucas and Fibonacci are identical.
            Is the Golden String involved ? (ie.1011010110110...)


            Maybe I've gone to far !
            Shane F.
          • djbroadhurst
            ... I think not, Shane. It seems to me that you have stayed in the same place, since F(2*n)=L(n)*F(n). David Broadhurst
            Message 5 of 12 , Jan 5, 2002
              Shane Findley wrote:
              > Maybe I've gone to far !
              I think not, Shane.
              It seems to me that you have stayed in the same place,
              since F(2*n)=L(n)*F(n).
              David Broadhurst
            • ttpi314159
              One more, issue. Case 2, seems to have 2^p +1 /3 as a prime divisor too ?
              Message 6 of 12 , Jan 5, 2002
                One more, issue.
                Case 2, seems to have 2^p +1 /3 as a prime divisor too ?
              • mikeoakes2@aol.com
                ... I guess what you mean is:- if L(2^(p-1)-1) mod (2^p-1) = 0 [CASE2} then L(2^(p-1)-1) mod (2^p+1)/3 = 0. Well, as per my previous communication, write q =
                Message 7 of 12 , Jan 6, 2002
                  Shane wrote:
                  > One more, issue.
                  > Case 2, seems to have 2^p +1 /3 as a prime divisor too ?

                  I guess what you mean is:-
                  if L(2^(p-1)-1) mod (2^p-1) = 0 [CASE2}
                  then L(2^(p-1)-1) mod (2^p+1)/3 = 0.

                  Well, as per my previous communication, write q = 2^p-1.

                  If CASE2, then q mod 5 = +-1, and q divides L(m) with m=(q-1)/2 = 2^(p-1)-1.
                  But q mod 5 = +1 => (2^p+1) mod 5 = 3 => (2^p+1)/3 mod 5 = +1,
                  and q mod 5 = -1 => (2^p) mod 5 = 0 which is impossible.
                  So in CASE2, (2^p+1)/3 mod 5 = 1,
                  so (2^p+1)/3 divides L(n) with n=((2^p+1)/3-1)/2 = (2^(p-1)-1)/3.
                  But any prime dividing L(n) also divides L(3*n)=L(2^(p-1)-1).
                  Q.E.D.

                  Mike Oakes
                • mikeoakes2@aol.com
                  In a message dated 05/01/2002 12:42:45 GMT Standard Time, ... I have obtained a partial answer to this question: there is no such Mersenne with (prime or
                  Message 8 of 12 , Jan 8, 2002
                    In a message dated 05/01/2002 12:42:45 GMT Standard Time,
                    d.broadhurst@... writes:

                    > Thanks for the clarification, Mike.
                    > But Shane's question was not answered by either of us:
                    > is there a Lucas mersenne pseudoprime,
                    > M(p)=2^p-1 with prime p and composite M(p)
                    > and one of the two tests satisfied?
                    > It's a nice question!
                    >

                    I have obtained a partial answer to this question: there is no such Mersenne
                    with (prime or composite) p <= 4500. [I'm sure you'll appreciate the number
                    of CPU cycles that went into this result, David -- L(2^4499) is quite big! --
                    did I compute it? -- "Were there but world enough and time", as the poet put
                    it...]

                    However, I expect there actually to be infinitely many such Mersennes, for
                    the following reason:- about one in 100,000 integers are "Lucas pseudoprimes"
                    (as defined in my previous posting), the first being 15251=101*151, the 23rd
                    being 1970299=199*9901 - and no, they don't ALL have just 2 primes factors:-).
                    So, if there is nothing special about the form (2^n-1) (a big IF), and if the
                    density of these pseudoprimes doesn't decrease too much with increasing size
                    (another big IF), then one would expect about 1 in 100,000 Mersennes to be
                    Lucas pseudoprimes.

                    Anyone up to shedding light on these IFs, and/or extending the search range
                    above p=4500?

                    Mike Oakes
                  • Shane
                    mikeoakes2@a... wrote: 05/01/2002 ... Mersenne with (prime or composite) p
                    Message 9 of 12 , Sep 30, 2002
                      mikeoakes2@a... wrote:
                      05/01/2002
                      > d.broadhurst@o... writes:
                      >
                      > > Thanks for the clarification, Mike.
                      > > But Shane's question was not answered by either of us:
                      > > is there a Lucas mersenne pseudoprime,
                      > > M(p)=2^p-1 with prime p and composite M(p)
                      > > and one of the two tests satisfied?
                      > > It's a nice question!
                      > >
                      > I have obtained a partial answer to this question: there is no such
                      Mersenne with (prime or composite) p <= 4500. [I'm sure you'll
                      appreciate the number of CPU cycles that went into this result,
                      David -- L(2^4499) is quite big! -- did I compute it? -- "Were
                      there but world enough and time", as the poet put it...]
                      > However, I expect there actually to be infinitely many such
                      Mersennes, for the following reason:- about one in 100,000 integers
                      are "Lucas pseudoprimes" (as defined in my previous posting), the
                      first being 15251=101*151, the 23rd
                      > being 1970299=199*9901 - and no, they don't ALL have just 2 primes
                      factors:-). So, if there is nothing special about the form (2^n-1)
                      (a big IF), and if the density of these pseudoprimes doesn't
                      decrease too much with increasing size (another big IF), then one
                      would expect about 1 in 100,000 Mersennes to be Lucas pseudoprimes.
                      > Anyone up to shedding light on these IFs, and/or extending the
                      search range
                      > above p=4500?
                      > Mike Oakes




                      Hello Mike, You had sent me a program for this, could you send it
                      again?

                      I am wondering if PRP/Newpen can be verified first by:
                      L(2^n-1) mod (k*2^n +/-1)=0
                      Then if positive execute PRP, and finally proth.

                      Does the 1/100,000 probability still hold?
                      What do you think about this variation?
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