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25310Re: What if Riemann's prime-counting formula was not the best?

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  • Chroma
    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
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