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

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• 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 1 of 16 , Jul 30, 2013
>
> [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
• ... 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 2 of 16 , Jul 30, 2013
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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