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Re: 2 > exp(Euler)

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  • djbroadhurst
    ... Hans Riesel discusses this on pp 66-67 of his book, following Theorem 3.1. He asserts that http://tinyurl.com/y9whej4 ... by a factor 2/exp(Euler) =~
    Message 1 of 22 , Sep 24, 2009
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > This reminds me of an old chestnut:
      > should one replace the Mertens constant
      > exp(Euler) =~ 1.78107
      > by 2, in this case, where N >> p >> 1.

      Hans Riesel discusses this on pp 66-67 of his book,
      following Theorem 3.1. He asserts that

      http://tinyurl.com/y9whej4

      > the sieve of Eratosthenes sieves out numbers
      > more efficiently than does a "random" sieve

      by a factor 2/exp(Euler) =~ 1.123.

      It appears that, as we approach p = sqrt(N), divisibility by
      primes is not statistically independent. However this is not
      explained further. It is merely inferred from comparing
      Mertens' theorem with the PNT.

      No wonder Tschebycheff was worried...

      David
    • mikeoakes2
      ... That s a nice follow-on, to consider primN(x,y,n). You have looked at my Case (5). I have devoted a couple of cpu days to investigating Case (2). For
      Message 2 of 22 , Sep 25, 2009
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        --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
        >
        > Here are a couple of gigantic PRPs:
        >
        > (101^5329+35^5329-lucasV(135,3535,5329))/
        > (101^73+35^73-lucasV(135,3535,73)) is Fermat and Lucas PRP!
        >
        > (106^7921+35^7921-lucasV(140,3710,7921))/
        > (106^89+35^89-lucasV(140,3710,89)) is Fermat and Lucas PRP!
        >
        > with 10535 and 15863 digits.
        >
        > I chose the composite index n = q^2 to be the
        > square of a prime so as to have only two divisors
        > of n in the Moebius transformation. For semiprime
        > n = q*r, there would be three:
        > primN(x,y,q*r) = N(x,y,q*r)/(N(x,y,q)*N(x,y,r))

        That's a nice follow-on, to consider primN(x,y,n).

        You have looked at my Case (5).
        I have devoted a couple of cpu days to investigating Case (2).

        For semiprime n, I have found several gigantic PRPs, for example:
        lucasV(1,-1,421*263)/(lucasV(1,-1,421)*lucasV(1,-1,263)),
        which is the same as L(421*263)/(L(421)*L(263)),
        with 22997 digits, and
        lucasV(1,-4,269*109)/(lucasV(1,-4,269)*lucasV(1,-4,109)) with 11824 digits.

        For n=q^2 the search space is more restricted, and I have been unable to find a gigantic PRP. The largest I can find is
        lucasV(1,-480^2,37^2)/lucasV(1,-480^2,37)
        with 3574 digits.

        Mike
      • djbroadhurst
        ... That was found by Bouk, 5 years ago: http://tinyurl.com/y9jos2d Please take care not to duplicate it, chez Henri. These are known PRPs that are gigantic
        Message 3 of 22 , Sep 25, 2009
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          --- In primenumbers@yahoogroups.com,
          "mikeoakes2" <mikeoakes2@...> wrote:

          > lucasV(1,-1,421*263)/(lucasV(1,-1,421)*lucasV(1,-1,263))

          That was found by Bouk, 5 years ago:
          http://tinyurl.com/y9jos2d

          Please take care not to duplicate it, chez Henri.

          These are known PRPs that are
          gigantic Lucas primitive parts
          with odd semi-prime indices:

          [ p*q [p , q] digits of primV(1,-1,p*q)]

          [ 83481, [3, 27827], 11631]
          [110723, [263, 421], 22997] \\ your rediscovery
          [138989, [23, 6043], 27780]
          [154281, [3, 51427], 21495]
          [177247, [7, 25321], 31750]
          [180309, [3, 60103], 25121]
          [184971, [3, 61657], 25771]
          [218341, [29, 7529], 44052]
          [220567, [367, 601], 45894]

          The biggest was found by Michael Porter in 2007.

          David
        • mikeoakes2
          ... Success at last: lucasV(1,-470^2,109^2)/lucasV(1,-470^2,109) has 31463 digits, and has been submitted to Henri s repository. Mike
          Message 4 of 22 , Sep 25, 2009
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            >
            > For n=q^2 the search space is more restricted, and I have been
            > unable to find a gigantic PRP.

            Success at last:
            lucasV(1,-470^2,109^2)/lucasV(1,-470^2,109) has 31463 digits,
            and has been submitted to Henri's repository.

            Mike
          • mikeoakes2
            ... There was one remaining inaccuracy here: I only gave the sieving depth to a rounded value of 50 billion. I have just done another NewPGen sieve and
            Message 5 of 22 , Sep 26, 2009
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              --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
              >
              > You were within 2 sigma of theory:
              >
              > \p6
              > pr109=1;forprime(q=2,109,pr109*=q);km=2*10^9;p=5*10^10;
              > tried=393614016;found=140719122;hitrate=1.*found/tried;
              > theory=exp(Euler)*log(p)*intnum(k=1,km,1/log(k*pr109))/km;
              > test=hitrate/theory;sigm=(test-1)*sqrt(found);
              > print([hitrate,theory,test,sigm])
              >
              > [0.357505, 0.357450, 1.00015, 1.82364]
              >
              > The fidelity with theory is now wonderful.
              >
              > Your AP15 project measured exp(Euler) to better
              > than 2 parts in 10^4, in 3 GHz-days.

              There was one remaining inaccuracy here: I only gave the sieving depth to a rounded value of 50 billion.

              I have just done another NewPGen sieve and recorded the exact depth. After pfgw'ing (combined processing time < 1 GHz-day), the figures are very satisfactory, as sigma is down from 1.82 to 1.53:-

              \p6
              pr83=1;forprime(q=2,83,pr83*=q);km=2*10^9;p=50066810303;
              tried=370597641;found=171068264;hitrate=1.*found/tried;
              theory=exp(Euler)*log(p)*intnum(k=1,km,1/log(k*pr83))/km;
              test=hitrate/theory;sigm=(test-1)*sqrt(found);
              print([hitrate,theory,test,sigm])

              [0.461601, 0.461547, 1.00012, 1.52844]

              Mike
            • djbroadhurst
              ... Great stuff, Mike. There is a corollary: both NewPgen and OpenPFGW must have been behaving near perfectly to give such wondrous agreement with Mertens,
              Message 6 of 22 , Sep 26, 2009
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                --- In primenumbers@yahoogroups.com,
                "mikeoakes2" <mikeoakes2@...> wrote:

                > [0.461601, 0.461547, 1.00012, 1.52844]

                Great stuff, Mike.

                There is a corollary: both NewPgen and OpenPFGW
                must have been behaving near perfectly to give
                such wondrous agreement with Mertens, nicht wahr?

                PS: How long will Pascal take for the AP16 :-?

                David
              • mikeoakes2
                ... Dang.., I hoped no-one (at least not one of my APn competitors:-) would spot what that NewPGen run was for; but you of course got it in one. On an Intel
                Message 7 of 22 , Sep 26, 2009
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                  --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
                  >
                  > PS: How long will Pascal take for the AP16 :-?

                  Dang.., I hoped no-one (at least not one of my APn competitors:-) would spot what that NewPGen run was for; but you of course got it in one.

                  On an Intel Quad at 3.6Ghz, about 4 months elapsed if I'm lucky, twice that if not.

                  (Current status: after 10 elapsed hours, it has found 7 AP13's.)

                  Mike
                • djbroadhurst
                  ... Oh what a tangled web remains, When author candidly explains! David, pace Marmion http://www.quotationspage.com/quote/27150.html
                  Message 8 of 22 , Sep 26, 2009
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                    --- In primenumbers@yahoogroups.com,
                    "mikeoakes2" <mikeoakes2@...> wrote:

                    > Dang.., I hoped no-one
                    > (at least not one of my APn competitors:-)
                    > would spot what that NewPGen run was for;
                    > but you of course got it in one.

                    Oh what a tangled web remains,
                    When author candidly explains!

                    David, pace Marmion
                    http://www.quotationspage.com/quote/27150.html
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