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

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  • Chris De Corte
    Alan, I said in my document that I only made a calculation up to 49978001 because I don t have other data available. If you all agree on a new trial then
    Message 1 of 16 , Jul 28, 2013
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      Alan,

      I said in my document that I only made a calculation up to 49978001 because I don't have other data available.

      If you all agree on a new trial then 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.

      Best,
      Chris


      ________________________________
      From: Alan Powell <AlanPowell@...>
      To: primenumbers@yahoogroups.com
      Cc: Chris De Corte <chrisdecorte@...>
      Sent: Sunday, July 28, 2013 1:42 PM
      Subject: Re: What if Riemann's prime-counting formula was not the best?



      Chris
       
      In the Wikipedia article titled “Prime-counting function”
       
        http://en.wikipedia.org/wiki/Prime-counting_function%c2%a0
       
      the value of Pi(10^24) is exactly 18,435,599,767,349,200,867,866.
       
      Your “preferred formula”
       
        Pi(x) = alpha*x^beta  with alpha=0.2083666 and beta=0.9294465
      gives
       
      only 4,222,251,563,919,881,535,488  which is out by a
      factor of 4+ !!
       
        You may want to study some elementary books on Number Theory to
      understand why Riemann’s function gives the best asymptotic value
      as the number of primes tends towards infinity.
       
      Regards
       
      Alan

      [Non-text portions of this message have been removed]
    • 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 2 of 16 , Jul 28, 2013
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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 3 of 16 , Jul 28, 2013
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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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          [Non-text portions of this message have been removed]
        • 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 4 of 16 , Jul 28, 2013
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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 5 of 16 , Jul 28, 2013
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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 6 of 16 , Jul 30, 2013
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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 7 of 16 , Jul 30, 2013
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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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