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Re: Prime Question

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  • jfoug@kdsi.net
    ... be ... the ... are ... stands ... Proth/NewPGen ... I ... 10000 ... Paul, I created a simple file in the new ABC2 format (usable by PFGW development
    Message 1 of 5 , Feb 1, 2001
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      --- In primenumbers@y..., paulmillscv@y... wrote:
      > Hi to all,
      > By popular request, a new Yahoo Groups number. This one could
      be
      > BIG.
      > It is 3*2^n –1 n is prime. Note a certain similarity to
      the
      > one and only 2^n –1 (Mersenne). Note also the 3,2,1. So here
      are
      > the first 3Mn numbers of note.
      >
      > (To start the first one for any n is 3*2^18 +- 1 A twin pair!)
      > Then the first 3Mn is 3*2^43 –1 then n=103 and the record
      stands
      > at n=827
      > Yes, 3*2^827 –1 is prime.
      >
      > Let's give a big hand to Proth and NewPGen courtesy of Gallot and
      > Jobling. (I knew it had to be easier to find titans) If you can't
      > beat them join them. So here are my first 2 titans
      > 3*2^4204 –1 (1267 digits) and 3*2^5134 – 1. (1546 digits)
      >
      > Not an ounce of theory so far!
      > Also, can we stand back and appreciate just what Proth and NewPGen
      > have accomplished? In a few seconds (OK, 10 seconds)
      Proth/NewPGen
      > have found a 3Mn (n = 827) which is not a million miles away from
      > Mn . It took all of us until 1952 (Robinson) to find 2^607 – 1.
      I
      > mean, this is PROGRESS! I think I can wait another 10 years and
      > type in 2^10^9 – 1 to find (is PRIME!?) in just a few seconds. I
      > can't wait!
      >
      > OK, so here is the puzzle of the moment. Can anyone improve on
      > n=827 n must be prime. I haven't checked the range n= 5000 –
      10000
      > so go for it!

      Paul,

      I created a simple file in the new ABC2 format (usable by PFGW
      development version). The file "aaa" is simply

      ABC2 3*2^$a-1
      a: primes from 2 to 10000

      I ran this using the command pfgw -f aaa

      This took a couple of minutes to run. The -f adds trial factoring to
      a pfgw search, which speeds up this unfactored search. It processed
      all numbers which n was prime from 2 to 10000 and these numbers came
      up as 3Mn primes (where n is prime):

      3*2^43-1
      3*2^103-1
      3*2^827-1
      3*2^7559-1

      So 7559 is the next prime satisfying the 3*2^n-1 with n prime.

      Note pfgw only did a PRP check when used as above. I still had to
      do a -tp primality test to validate that this number was prime:
      pfgw -tp "-q3*2^7559-1"

      pfgw can be found in the primeform egroup (or now Yahoo group).
      http://groups.yahoo.com/group/primeform
      You might want to add this program to your prime searching arsenal.

      Jim.
    • Yves Gallot
      ... ... and few weeks later, we obtain: rank description digits who year ... 212 3*2^164987-1 49667 gb 1999 266 3*2^155930-1
      Message 2 of 5 , Feb 1, 2001
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        > 3*2^43-1
        > 3*2^103-1
        > 3*2^827-1
        > 3*2^7559-1
        >
        > So 7559 is the next prime satisfying the 3*2^n-1 with n prime.

        ... and few weeks later, we obtain:

        rank description digits who year
        ---- -------------------- ------ --- ----
        212 3*2^164987-1 49667 gb 1999
        266 3*2^155930-1 46941 gb 1999
        568 3*2^123630-1 37217 gb 1999
        1153 3*2^97063-1 29220 gb 1999
        1373 3*2^88171-1 26543 gb 1999
        1507 3*2^85687-1 25795 gb 1998
        1797 3*2^80330-1 24183 gb 1998
        2746 3*2^71783-1 21610 gb 1998

        Yves
      • paulmillscv@yahoo.co.uk
        Hi to all, Many thanks for the info about the 3Mn number and (aMn numbers) from various people. Yves Gallot points out that the record for 3Mn stands at
        Message 3 of 5 , Feb 5, 2001
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          Hi to all,
          Many thanks for the info about the 3Mn number and (aMn
          numbers) from various people. Yves Gallot points out that the record
          for 3Mn stands at 3*2^164987 -1 is prime. 164987 is prime!

          Not one to stray from theory too long I think I have come up
          with a way to use the Proth/NewPGen combo to arrive at an interesting
          equation/conjecture. The equation is

          n*X = p + 1 n is an integer, n>= 1. X is an integer >=1
          and p is any prime.

          Some simple solutions are
          n= 1 X= 3 p = 2
          n= 2 X= 2 p = 3
          n= 3 X= 1 p = 3
          n= 4 X= 1 p = 3
          n =5 X= 4 p = 19 etc

          We see that if X = 1 then n = p + 1 so we can write down a solution
          for every integer n equal to a prime value p plus 1. I.e the
          equation has a solution for n = 3,4,6,8,12 etc (all p + 1) . Note
          how the minimal solution for n= 5 jumped to p = 19 from p = 3 so this
          is not a trival series.
          I will be working on solutions to this equation for n <= 100,000 at
          first, to see how we get on. A minimal solution to this equation for
          any n will be the smallest X and p. The use of this equation will be
          evident because it connects the 'hats off, pause' integers n, with
          primes. So if anyone wants to leap in and start finding solutions,
          go ahead. I will compile the results. Note that solutions to this
          equation can be enjoyably found with Proth/NewPGen set to values such
          as A*2^25 - 1 = p and just listing the values of A which give the 'is
          prime!' value. Then A*2^25 = p + 1 and we have a solution for n = A
          which is X = 2^25 and p is some value. The point is that we leave
          the prime number as part of the equation so we don't have to know its
          decimal expansion.
          Sounds fun? Good luck!

          regards,
          Paul Mills
          Kenilworth, England
        • asen@theory.saha.ernet.in
          ... Now,if n=m what do we have? 2^n=1+n*p This implies n can never be even...except of course n=0...then it is trivial. If n is not even...cann t say offhand!
          Message 4 of 5 , May 30, 2001
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            > > Hi!
            > > First of all...the two n's on either side do not seem to be the same.So
            > > you basically say
            > > 2^n-1=m*p.Right?We'll say about n=m in a while.
            > > Now it seems trivial.
            > > 2^n-1 is odd and we know any odd number is factorisable into odd integers.
            > > Now taking one of the prime factors as p and the rest as m the result
            > > follows.
            > > Did I clear the point or did I miss it altogether?
            > > Again,for Mersenne primes m=1.
            Now,if n=m what do we have?
            2^n=1+n*p
            This implies n can never be even...except of course n=0...then it is
            trivial.
            If n is not even...cann't say offhand!


            > >
            > > On Wed, 30 May 2001 paulmillscv@... wrote:
            > >
            > > > Hi to all,
            > > > Apologies for not being on the list recently but the local TV
            > > > station did a rerun of the Pink Panther movies. So, I wish to
            > > > have "speaks" with the group.
            > > >
            > > > I have 'good reason to believe' that
            > > > 2^n - 1 = n*p for some integer n, p a prime.
            > > >
            > > > n is odd, "I know that, I know that.."
            > > >
            > > > Can you prove me wrong, right!
            > > >
            > > > regards
            > > > Paul Mills
            > > > Keniworth,
            > > > England.
            > > >
            > > >
            > > > Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
            > > > The Prime Pages : http://www.primepages.org
            > > >
            > > >
            > > >
            > > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
            > > >
            > > >
            > >
            > >
            >
            >
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