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Re: 2*a^a+-1 primes

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  • Paul Underwood
    ... available ... The look of your pages look familiar; lol. I got my colours from mersenneforum.org I use a yahoo database to store my table; then export it;
    Message 1 of 11 , Sep 1, 2003
      --- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
      > HI,
      > I am starting a new distributed computing project which is
      available
      > at
      >
      > http://groups.yahoo.com/groups/primenumbers/files/2*a^a+-1 prime
      > search/index.htm
      >

      The look of your pages look familiar; lol.

      I got my colours from mersenneforum.org

      I use a yahoo database to store my table; then export it; run a perl
      script over the export to produce the html -- this is not very smart!
      Hopefully you won't have to do much editing of your status table html.

      Good luck with the distributed search

      Paul
    • mikeoakes2@aol.com
      ... The how long does it take heading on that page says:- ... Remember: this is not true for the larger values of a , as each test takes time proportional
      Message 2 of 11 , Sep 2, 2003
        --- In primenumbers@yahoogroups.com, harsh@... (eharsh82) wrote:
        > I am starting a new distributed computing project which is available
        > at
        >
        > http://groups.yahoo.com/groups/primenumbers/files/2*a^a+-1 prime
        > search/index.htm
        >

        The "how long does it take" heading on that page says:-
        > About 10000 candidates can be tested on a P900 every week.

        Remember: this is not true for the larger values of "a", as each test takes
        time proportional to approximately (a*log(a))^2, and so increases quite
        rapidly with increasing a.

        So people should not reserve /too/ large ranges!

        Mike



        [Non-text portions of this message have been removed]
      • mikeoakes2@aol.com
        ... As part of my investigation of the factor structure of these numbers, I have run PFGW with -d -f200 on all values of a
        Message 3 of 11 , Sep 3, 2003
          In a message dated 31/08/03 17:28:26 GMT Daylight Time, harsh@... writes:


          > all the known primes of this type
          >
          > 2*(1^1)+1
          > 2*(12^12)+1
          > 2*(18^18)+1
          > 2*(251^251)+1
          >
          >
          > 2*(2^2)-1
          > 2*(3^3)-1
          > 2*(357^357)-1
          > 2*(1400^1400)-1
          >
          > all these were found by me
          >
          > Harsh Aggarwal
          >

          As part of my investigation of the factor structure of these numbers, I have
          run PFGW with "-d -f200" on all values of a <= 5000, for both forms, which
          finds all factors up to about 1000000, and took about 4 Athlon-GHz-days.

          This confirmed Harsh's results above.
          And it also uncovered quite a few big PRPs.
          The following 3, only, are larger than 10000 digits and so eligible for Henri
          Lifchitz's "Top 5000 PRP" database:-

          (2*2959^2959+1)/(3*14753*148711)
          (2*3214^3214+1)/(3*11*6427)
          (2*4701^4701+1)/(17*29^2*4703*18803*122869)

          Mike


          [Non-text portions of this message have been removed]
        • mikeoakes2@aol.com
          How many primes N=2*a^a+-1 can we expect for X_0
          Message 4 of 11 , Sep 3, 2003
            How many primes N=2*a^a+-1 can we expect for X_0 <= a <= X?

            If N was a "typical" integer of that size, the probability of its being prime
            would be 1/log(N), by the PNT.

            But N is clearly not divisible by 2, so we must multiply by 2.

            So, ignoring the complicated story of which primes can divide which of these
            numbers (see another of my posts), which can be expected to add only an
            overall multiplying factor O(1), a first guess would be:-
            count = integral{X_0 to X}dx/log(N) = integral{X_0 to X}dx/(x*log(x)+log(2))
            ~ integral{X_0 to X}d(log(x))/log(x) ~ [log(log(X))] as X => infinity.

            So first of all: this heuristic argument indicates that there are indeed
            infinitely many primes of each type.

            Putting in the numbers gives the following table:-
            X log(X) count=log(log(X))
            10 2.3026 0.834
            10^2 4.6052 1.527
            10^3 6.9078 1.933
            10^4 9.2103 2.220
            10^5 11.5129 2.443
            10^6 13.8155 2.626
            10^7 16.1181 2.780
            10^8 18.4207 2.913
            10^9 20.7233 3.031
            10^10 23.0259 3.137

            The good news: for both the "+" and "-" formulae, the number of known primes
            (a <= 5*10^3) is actually 4, which gives some confidence that the heuristics
            are on the right track, given that PNT is only very approximate for these small
            values of X, etc :-)

            The bad news: the probability for finding a new prime increases _extremely_
            slowly with X :-(

            Mike Oakes


            [Non-text portions of this message have been removed]
          • eharsh82
            I expect a prime for the + series under 35000 and for the - series under 30000 Harsh Aggarwal ... being prime ... of these ... only an ... +log(2)) ...
            Message 5 of 11 , Sep 3, 2003
              I expect a prime for the + series under 35000
              and for the - series under 30000

              Harsh Aggarwal

              --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
              > How many primes N=2*a^a+-1 can we expect for X_0 <= a <= X?
              >
              > If N was a "typical" integer of that size, the probability of its
              being prime
              > would be 1/log(N), by the PNT.
              >
              > But N is clearly not divisible by 2, so we must multiply by 2.
              >
              > So, ignoring the complicated story of which primes can divide which
              of these
              > numbers (see another of my posts), which can be expected to add
              only an
              > overall multiplying factor O(1), a first guess would be:-
              > count = integral{X_0 to X}dx/log(N) = integral{X_0 to X}dx/(x*log(x)
              +log(2))
              > ~ integral{X_0 to X}d(log(x))/log(x) ~ [log(log(X))] as X =>
              infinity.
              >
              > So first of all: this heuristic argument indicates that there are
              indeed
              > infinitely many primes of each type.
              >
              > Putting in the numbers gives the following table:-
              > X log(X) count=log(log(X))
              > 10 2.3026 0.834
              > 10^2 4.6052 1.527
              > 10^3 6.9078 1.933
              > 10^4 9.2103 2.220
              > 10^5 11.5129 2.443
              > 10^6 13.8155 2.626
              > 10^7 16.1181 2.780
              > 10^8 18.4207 2.913
              > 10^9 20.7233 3.031
              > 10^10 23.0259 3.137
              >
              > The good news: for both the "+" and "-" formulae, the number of
              known primes
              > (a <= 5*10^3) is actually 4, which gives some confidence that the
              heuristics
              > are on the right track, given that PNT is only very approximate for
              these small
              > values of X, etc :-)
              >
              > The bad news: the probability for finding a new prime increases
              _extremely_
              > slowly with X :-(
              >
              > Mike Oakes
              >
              >
              > [Non-text portions of this message have been removed]
            • mikeoakes2@aol.com
              ... On what grounds? (please enlighten us;-) Mike [Non-text portions of this message have been removed]
              Message 6 of 11 , Sep 4, 2003
                In a message dated 04/09/03 02:03:15 GMT Daylight Time, harsh@... writes:


                > I expect a prime for the + series under 35000
                > and for the - series under 30000
                >

                On what grounds?
                (please enlighten us;-)

                Mike


                [Non-text portions of this message have been removed]
              • Paul Jobling
                Mike, You said a while back that you had a proof regarding the divisiblity of properties of the primes=1 mod 4 with these numbers. Could you post it here for
                Message 7 of 11 , Sep 4, 2003
                  Mike,

                  You said a while back that you had a proof regarding the divisiblity of
                  properties of the primes=1 mod 4 with these numbers. Could you post it here
                  for us at all? I for one would certainly be interested.

                  Regards,

                  Paul.



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                • mikeoakes2@aol.com
                  In a message dated 04/09/03 09:17:45 GMT Daylight Time, ... So there is one reader out there who is interested, after all! Let p be a prime = +1 mod 4, and N =
                  Message 8 of 11 , Sep 4, 2003
                    In a message dated 04/09/03 09:17:45 GMT Daylight Time,
                    Paul.Jobling@... writes:


                    > You said a while back that you had a proof regarding the divisiblity of
                    > properties of the primes=1 mod 4 with these numbers. Could you post it here
                    > for us at all? I for one would certainly be interested.
                    >

                    So there is one reader out there who is interested, after all!

                    Let p be a prime = +1 mod 4, and N = 2*a^a+1.
                    Let a = s mod p, 0 <= s <= p-1,
                    and a = t mod (p-1), 0 <= t <= p-2.

                    N = 0 mod p iff
                    2*a^a = -1 mod p,
                    i.e. a^a = (p-1)/2 mod p,
                    i.e. s^a = (p-1)/2 mod p,
                    i.e. s^t = (p-1)/2 mod p, since s^(p-1) = 1 mod p (Fermat).

                    So, for each t in [0,p-2] we are looking for solutions s in [0,p-1] of
                    s^t = (p-1)/2 mod p (1)

                    Repeating the above for the "-" case, we are looking for solutions of
                    s^t = -(p-1)/2 mod p (2)

                    If t is odd, each solution of (1) gives a solution of (2) (just change the
                    sign of s), and vice versa.
                    If t is even, s^2 = -1 has a solution (-1 is a quadratic residue for primes p
                    = +1 mod 4), so again, each solution of (1) gives a solution of (2), and vice
                    versa.

                    QED

                    Mike


                    [Non-text portions of this message have been removed]
                  • mikeoakes2@aol.com
                    As Harsh announced here a while back, there is a small coordinated search going on for primes of this form - see:
                    Message 9 of 11 , Sep 16, 2003
                      As Harsh announced here a while back, there is a small coordinated search
                      going on for primes of this form - see:
                      http://groups.yahoo.com/group/primenumbers/files/2%2Aa%5Ea%2B-1%20prime%20sear
                      ch/index.htm

                      so it is instructive to see how much more or less likely such numbers are to
                      be prime than "random" integers of the same size.

                      Here are the results of removing all prime factors < p_max for both the "-"
                      and "+" forms, compared with the same exercise for a set of numbers of the form
                      2*a^a+r, where the "r" values are selected randomly from the range 1<=r<=10^9.

                      A range of 20000 <= a <= 25000 was used, for all 3 cases, and the columns in
                      the following table record how many of the 5000 starting numbers remain after
                      sieving to the depth p_max:-

                      p_max "-" form "+" form "random" form
                      2*10 2734 1650 1698
                      55 2286 1367 1365
                      2*10^2 1856 1167 1198
                      2*10^3 1395 861 829
                      2*10^4 1020 675 642
                      2*10^5 592 318 479
                      2*10^6 501 247 420
                      2*10^7 431 205 ??
                      2*10^8 381 178 ??
                      2*10^9 351 156 ??

                      Phil Carmody's excellent "aagenm.exe" was used to obtain the results in the
                      2nd and 3rd columns, in about 0.8GHz-days.
                      For the final column PFGW was used, but its trial division is about 100 times
                      slower than Phil's program's, and so would have taken months to find all the
                      ?? figures.

                      It is clear that more primes are to be expected from the "-" form than either
                      from the "+" form or from a collection of "ordinary" (odd) numbers of the
                      same size.

                      The p_max=55 row is included to enable a comparison to be made with the
                      theoretical prediction from my earlier posts giving the probabilities of the "-"
                      and "+" forms being divisible by each of the primes <= 53:-

                      p_max "-" form "+" form "random" form
                      55 2221.8 1419.8 1360.9

                      where the last column is of course just 5000*(1-1/3)*(1-1/5)*...*(1-1/53).

                      Mike


                      [Non-text portions of this message have been removed]
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