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Re: Lucas mersenne ?

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  • djbroadhurst
    ... 8191 divides L(4095)
    Message 1 of 12 , Jan 2, 2002
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      > 2^13-1 divide L(4095,4096) ?
      8191 divides L(4095)
    • djbroadhurst
      ... p CASE [ 2, CASE1] [ 3, CASE1] [ 5, CASE2] [ 7, CASE1] [13, CASE2] [17, CASE2] [19, CASE1] interesting....
      Message 2 of 12 , Jan 2, 2002
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        > 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....
      • 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 3 of 12 , Jan 2, 2002
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          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 4 of 12 , Jan 5, 2002
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            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 5 of 12 , Jan 5, 2002
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              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 6 of 12 , Jan 5, 2002
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                --- 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 7 of 12 , Jan 5, 2002
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                  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 8 of 12 , Jan 5, 2002
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                    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 9 of 12 , Jan 6, 2002
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                      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 10 of 12 , Jan 8, 2002
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                        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 11 of 12 , Sep 30, 2002
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                          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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