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Re: What if Riemann's prime-counting formula was not the best?

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  • djbroadhurst
    ... Please use plain text in messages to this list, else you may not be understood. Here I provide [N, pi(10^N), R(10^N)] [10, 455052511, 455050683] [11,
    Message 1 of 16 , Jul 28 12:11 PM
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      --- In primenumbers@yahoogroups.com,
      Chris De Corte <chrisdecorte@...> wrote:

      > please provide me with 10 N, 10 pi(x) and  10 best Riemann's
      > approximations and I will try to calculate a new alpha and beta.

      Please use plain text in messages to this list, else
      you may not be understood. Here I provide

      [N, pi(10^N), R(10^N)]
      [10, 455052511, 455050683]
      [11, 4118054813, 4118052495]
      [12, 37607912018, 37607910542]
      [13, 346065536839, 346065531066]
      [14, 3204941750802, 3204941731602]
      [15, 29844570422669, 29844570495887]
      [16, 279238341033925, 279238341360977]
      [17, 2623557157654233, 2623557157055978]
      [18, 24739954287740860, 24739954284239494]
      [19, 234057667276344607, 234057667300228940]
      [20, 2220819602560918840, 2220819602556027015]
      [21, 21127269486018731928, 21127269485932299724]
      [22, 201467286689315906290, 201467286689188773625]
      [23, 1925320391606803968923, 1925320391607837268776]
      [24, 18435599767349200867866, 18435599767347541878147]
      [25, 176846309399143769411680, 176846309399141934626966]

      where

      R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

      David
    • Chris De Corte
      Hi, This is the result: N Pi(x) R(x) Power 10 1E+10 455052511 455050683 420213980 11 1E+11 4118054813 4118052495 3951369042 12 1E+12 37607912018 37607910542
      Message 2 of 16 , Jul 28 12:59 PM
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        Hi,

        This is the result:

        N Pi(x) R(x) Power
        10 1E+10 455052511 455050683 420213980
        11 1E+11 4118054813 4118052495 3951369042
        12 1E+12 37607912018 37607910542 37155635094
        13 1E+13 3.46066E+11 3.46066E+11 349383012475
        14 1E+14 3.20494E+12 3.20494E+12 3285329105477
        15 1E+15 2.98446E+13 2.98446E+13 30892707847616
        16 1E+16 2.79238E+14 2.79238E+14 290491262067780
        17 1E+17 2.62356E+15 2.62356E+15 2731556383920000
        18 1E+18 2.474E+16 2.474E+16 25685455133563200
        19 1E+19 2.34058E+17 2.34058E+17 241526262940073000
        20 1E+20 2.22082E+18 2.22082E+18 2271127195778980000
        21 1E+21 2.11273E+19 2.11273E+19 21355933208335000000
        22 1E+22 2.01467E+20 2.01467E+20 200814769003914000000
        23 1E+23 1.92532E+21 1.92532E+21 1888307621900430000000
        24 1E+24 1.84356E+22 1.84356E+22 17756192398666400000000
        25 1E+25 1.76846E+23 1.76846E+23 166965575334147000000000

        alpha 0.07774984570
        beta 0.9732770961
        correlation R(x) 1.00000000000
        correlation power 0.999997766
        So, yes, Riemann function is better in this test.
        I also attach the excel for those who can open it.

        I based my testing on a book I was reading about unsolved problems that wrote that Riemann had used the formula x/ln(x) which was obviously an oversimplification.

        I am sorry if I wasted your guys day.

        Thanks & good night,
        Chris



        ________________________________
        From: djbroadhurst <d.broadhurst@...>
        To: primenumbers@yahoogroups.com
        Sent: Sunday, July 28, 2013 9:11 PM
        Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?




        --- In primenumbers@yahoogroups.com,
        Chris De Corte <chrisdecorte@...> wrote:

        > please provide me with 10 N, 10 pi(x) and  10 best Riemann's
        > approximations and I will try to calculate a new alpha and beta.

        Please use plain text in messages to this list, else
        you may not be understood. Here I provide

        [N, pi(10^N), R(10^N)]
        [10, 455052511, 455050683]
        [11, 4118054813, 4118052495]
        [12, 37607912018, 37607910542]
        [13, 346065536839, 346065531066]
        [14, 3204941750802, 3204941731602]
        [15, 29844570422669, 29844570495887]
        [16, 279238341033925, 279238341360977]
        [17, 2623557157654233, 2623557157055978]
        [18, 24739954287740860, 24739954284239494]
        [19, 234057667276344607, 234057667300228940]
        [20, 2220819602560918840, 2220819602556027015]
        [21, 21127269486018731928, 21127269485932299724]
        [22, 201467286689315906290, 201467286689188773625]
        [23, 1925320391606803968923, 1925320391607837268776]
        [24, 18435599767349200867866, 18435599767347541878147]
        [25, 176846309399143769411680, 176846309399141934626966]

        where

        R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

        David



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      • djbroadhurst
        ... That was, of course, a log-lover s spoof on log-haters. However, this log-hater: http://arxiv.org/pdf/1307.4444.pdf attempts to fix up the obvious lunacy
        Message 3 of 16 , Jul 28 1:24 PM
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          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > Poly=polinterpolate(vector(25,k,10^k),data);

          That was, of course, a log-lover's spoof on log-haters.

          However, this log-hater:
          http://arxiv.org/pdf/1307.4444.pdf
          attempts to fix up the obvious lunacy of polynomial
          approximation of pi(x).

          Andrey Kulsha may offer us tighter conjectural
          bounds on pi(10^26) ?

          David
        • djbroadhurst
          ... On the assumption that Delta(x) defined by Andrey in http://www.primefan.ru/stuff/primes/table.html#theory continues to satisfy abs(Delta(x))
          Message 4 of 16 , Jul 28 8:21 PM
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            --- In primenumbers@yahoogroups.com,
            "djbroadhurst" <d.broadhurst@...> wrote:

            > Andrey Kulsha may offer us tighter conjectural
            > bounds on pi(10^26) ?

            On the assumption that Delta(x) defined by Andrey in
            http://www.primefan.ru/stuff/primes/table.html#theory
            continues to satisfy abs(Delta(x)) < 1, I estimate that

            pi(10^26) = 1699246750872419991992147 +/- 167036339194

            David (subject to error, as ever)
          • Chroma
            djbroadhurst wrote: ... For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms, Much faster can be calculated
            Message 5 of 16 , Jul 30 4:00 AM
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              djbroadhurst wrote:>
              >
              > [N, pi(10^N), R(10^N)]
              ........
              > [25, 176846309399143769411680, 176846309399141934626966]
              >
              > where
              >
              > R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
              >
              For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms,
              Much faster can be calculated as

              pi(x) ~= pli(x) = round(Li(x) - 1/2 Li(sqrt(x)))

              where Li(x) is the Logarithm integral

              pli(10^25) = 176846309399141938590795
              (pli(10^25)/R(10^25)) - 1 = 2 10^-17
              (pli(10^25)/pi(10^25)) -1 = -1 10^-14

              pli{10^250)= 1740206254656916846774941665048386410178028975968929264655269395003484\
              7365084787720410883002915274182213664956284195372937010842285191263145\
              7678993892420170619475710388189158537825404886895382231933346054713467\
              85875358018952542776800464839768387582

              --
              marian otremba
            • djbroadhurst
              ... No. The Gram formula is still very convenient at this size. Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:
              Message 6 of 16 , Jul 30 2:57 PM
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                --- In primenumbers@yahoogroups.com,
                Chroma <chromatella@...> wrote:

                >> R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
                > For large values of x, this algorithm is inconvenient,
                > eg for x = 10^250 requires over 1868 terms

                No. The Gram formula is still very convenient at this size.
                Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:

                R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

                {default(realprecision,260);print(R(10^250));
                print(" took "gettime" milliseconds");

                17402062546569168467749416650483864101780289759689292646552693950034847365084787720410883002915274182213664956284195372937010842285191263145767899389242017061947571038442681072462756632213511422607548574658029047365218974809766827365028215685475746
                took 98 milliseconds

                Perhaps you are paying for inferior software?
                If so, the general rule is: the less you pay,
                the better the deal.

                Pari-GP is totally free and hence rather hard to beat :-)

                David
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