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[PrimeNumbers] Re: Lucas mersenne ?

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  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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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