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Re: 2*a^a+-1 primes

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  • 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 1 of 11 , Sep 2, 2003
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      --- 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



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    • 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 2 of 11 , Sep 3, 2003
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        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


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      • mikeoakes2@aol.com
        How many primes N=2*a^a+-1 can we expect for X_0
        Message 3 of 11 , Sep 3, 2003
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          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


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        • 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 4 of 11 , Sep 3, 2003
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            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 5 of 11 , Sep 4, 2003
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              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


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            • 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 6 of 11 , Sep 4, 2003
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                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 7 of 11 , Sep 4, 2003
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                  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


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                • 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 8 of 11 , Sep 16, 2003
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                    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


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