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

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  • 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 1 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 2 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 3 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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