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a new way to proof Mersenne numbers , but faster ???

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  • Norman Luhn
    Hello members, years ago, I had have an idea to proof Mersenne numbers only with the exponent. So the largest number of M(45981737) we use is smaller than 93M.
    Message 1 of 1 , Jun 28, 2009
      Hello members,

      years ago, I had have an idea to proof Mersenne numbers only with the exponent. So the largest number of M(45981737) we use is smaller than
      93M.

      The problem, is it faster? I use also the Lucas-Lehmer Test.


      Here a confuse example:

      n=2^5-1 ,p=5

      start with 14: Dual = 2^3 + 2^2+2^1 :short 3 2 1

      Calculate 14^2 : we get matrix

      6 5 4, 5 4 3, 4 3 2 : optimize with p=5
      1 0 4, 0 4 3, 4 3 2 : so 4 + 4 =5 , I mean 2^4+2^4 =2^5 , 2^5 mod 2^5-1=1

      so 4+4=0 mod (p=5)

      after all we get : 3 1 -> 2^3+2^1=10 : 10-2=8 , here only 3 , because (14^2-2) = 8 mod 31

      3+3=6 : 6 mod 5 = 1 -> 2^6 mod 31 = 2 (2=2^"1"): 1-1=0 : Rest 0 , 2^5-1 is prime.


      Best

      Norman















      thanks andersen for your information (links).



      my algo supose that : for double prime mersenne :



      p = (2^n)-1 : is mersenne numbre :



      if (p - (n+1)) /n = prime then (2^p)-1 is probable double mersenne prime



      exempe : p = (2^5)-1



      = 31 is mersenne numbre



      test : (p-(n+1) / n



      = 31-(5+1)/5



      = 5 is prime so (2^p)-1 is probable prime = (2^31)-1



      exempe : p = (2^7)-1



      = 127 is mersenne numbre



      test : (127-(7+1) / 7



      = 17 is prime so (2^p)-1 is probable prime = (2^127)-1



      exempe : p = (2^13)-1



      = 8191 is mersenne numbre



      test : (p-(n+1) / n



      = 8191-(13+1)/ 13



      = 629 is not prime so (2^p)-1 is not probable prime = (2^8191)-1



      exempe : p = (2^17)-1



      = 131071 is mersenne numbre



      test : (p-(n+1) / n



      = 131071-(17+1) /17



      = 7709 is not prime so (2^p)-1 is not probable prime = (2^131071)-1



      ex .........



      for the mersenne numbre p = (2^521)-1



      my be (p-(n+1) / n is probable prime then



      2^ (2^521)-1 is a probable double mersenne prime if (p-(n+1)/n =prime



      thanks



      rachid baih



      > From: jens.k.a@get2net. dk

      > To: sta_staf@hotmail. com

      > Subject: Re: sorry

      > Date: Sat, 27 Jun 2009 01:16:10 +0200

      >

      > There are infinitely many composite n not of form 2^p-1 such that 2^n - 2

      > mod n = 0.

      > See for example http://www.research .att.com/ ~njas/sequences/ A001567

      > and even n in http://www.research .att.com/ ~njas/sequences/ A006935

      >

      > --

      > Jens Kruse Andersen (off list)

      >

      > ----- Original Message -----

      > From: "sta staf" <sta_staf@hotmail. com>

      > To: <primenumbers@ yahoogroups. com>

      > Sent: Friday, June 26, 2009 11:56 PM

      > Subject: [PrimeNumbers] sorry

      >

      >

      >

      > sory for all because the 2 numbre are not the mersenne numbre .

      > the great probleme is : my 2^n - 2 mod n = 0 if n is prime

      > this formula work with al integer else 2^p-1

      > example 2^p - 1 = 2^11 - 1 = 2047

      > 2^2047 - 2 mod 2047 = 0 but 2047 is not prime

      > so the formul 2^n - 2 mod n = 0 is corecte if n # 2p -1

      > for n = 99000313

      > 2^n-2 mod n = 0

      >



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