--- 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