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

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  • djbroadhurst
    ... I think that Chris considers log to be a complicated function. Here is a perfect fit to Jens data, without using the dreaded log :
    Message 1 of 16 , Jul 28, 2013
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      --- In primenumbers@yahoogroups.com,
      "Jens Kruse Andersen" <jens.k.a@...> wrote:

      > So let's make a comparison with exactly computed pi(x) values.
      > x/log(x) and x/(log(x)-1) are the commonly used approximations
      > not using li(x) or long formulas.

      I think that Chris considers "log" to be a complicated function.
      Here is a perfect fit to Jens' data, without using the dreaded "log":

      {data=[4,25,168,1229,9592,78498,664579,5761455,50847534,455052511,
      4118054813,37607912018,346065536839,3204941750802,29844570422669,
      279238341033925,2623557157654233,24739954287740860,234057667276344607,
      2220819602560918840,21127269486018731928,201467286689315906290,
      1925320391606803968923,18435599767349200867866,176846309399143769411680];}

      Poly=polinterpolate(vector(25,k,10^k),data);
      fit(n)=subst(Poly,x,n);

      for(k=1,25,print1(fit(10^k)-data[k]","));
      0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

      Looks good, to log-haters? Ought to give a good
      prediction to pi(10^26), surely :-?

      print(fit(10.^26));
      -6.2925202314581568492313228001263868142 E300

      Hmm.. Maybe that pesky log function has it uses?

      David
    • WarrenS
      Riemann s (& variant form) prime-counting formulas are the best in the sense they yield exact equality. Everywhere. Given various exact formulas and
      Message 2 of 16 , Jul 28, 2013
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        Riemann's (& variant form) prime-counting formulas are "the best" in the sense they
        yield exact equality. Everywhere.

        Given various exact formulas and algorithms for computing pi(x),
        the "best" would seem to be the one permitting the fastest computation (maybe memory
        consumption should also be taken into account).
        And in fact, the fastest currently known algorithm (asymptotically) for computing
        pi(x) is based on a form of Riemann, not combinatorial counting methods and not just generating primes. Also, for inexact computation of pi(x) to specified accuracy, this method still seems best known.

        However, there is no reason to believe that further speedups are necessarily impossible
        with further algorithmic/analytic ideas nobody has thought of yet.
        It would help if anybody had the faintest idea how to prove lower bounds on the
        computational complexity of counting primes.

        I could continue. With Green-Tao, we know there are arbitrarily long arithmetic progressions with all-prime entries. How computationally complex is it to find an N-term example? Well... it seems to be extremely hard... quite likely humanity will
        never find an example with N=50... but I am unaware of any
        reasonable lower bounds, or upper bounds, on this computational complexity!
      • 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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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