Bill Allombert on Sun, 27 Nov 2022 13:01:39 +0100 |
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Re: 3^k performance |
On Sat, Nov 26, 2022 at 08:47:39AM +0100, Ruud H.G. van Tol wrote: > > A022330_1(n)=1+n+sum(k=1,n,logint(3^k,2)) > > A022330_2(n)=my(t=1);1+n+sum(k=1,n,logint(t*=3,2)) > > ? A022330_1(10^5) > cpu time = 8,322 ms, real time = 8,352 ms. > %15 = 7924941755 > > ? A022330_2(10^5) > cpu time = 215 ms, real time = 217 ms. > %16 = 7924941755 > > > So about a 40x difference. > > > Is that worthwhile to look into? > How to best approach that? I suggest A022330_3(n)= { my(s=1+n,p3=1,p2=1,l2=1); for(k=1,n, p3 *= 3; p2 <<= 1; l2++; if(p3 > p2, p2 << = 1; l2++); s+=l2); } Cheers, Bill.