## Re: Conjecture Ludovicus V

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• ... Here, Shanks remarks that ... To get more than 500 decimal places of the HL constant, in less than 80 GHz-seconds, it suffices to consider primes up to
Message 1 of 6 , Nov 12, 2012
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"mikeoakes2" <mikeoakes2@...> wrote:

Here, Shanks remarks that

> With this improvement, the first two p's,
> i.e., 5 and 13, already suffice to yield
> the five decimal places shown.

To get more than 500 decimal places of the HL constant, in
less than 80 GHz-seconds, it suffices to consider primes up
to 31601 and to augment that data with the values of
beta(2*k) = suminf(n=0,(-1)^n/(2*n+1)^(2*k))
for k = 1 to 60. Note however that Pari-GP's "zetak" is
broken at this high precision, so one needs a few lines of
code, calling "incgam", for the summand of Proposition 3.2
of "High precision computation of Hardy-Littlewood
constants" by Henri Cohen, available as a .dvi file from
http://www.math.u-bordeaux1.fr/~hecohen/

David
• ... Solution: 141322 Method: Using incgam , eint1 and moebius this solution may be found in less than 8 seconds, running at less than 3 GHz:
Message 2 of 6 , Nov 17, 2012
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> Here are more than 500 good digits of the relevant
> Hardy-Littlewood constant
>
> 0.686406731409123004556096348363509434089166550627\
> 87977896811707366392111335868511586385990346954399\
> 18910971210114370836967066905693197374076733762228\
> 99676181590249072148452433271956678446776133507821\
> 66911238709121761546057993037867560421525052339604\
> 30362058008970065288339284308924109839850906244857\
> 05610181784829261874808914954658125072564596078857\
> 97178488488284913596339147096072562959833915384281\
> 95797768815771346844648575433020873782746429273861\
> 06852378388023207827786598975144050262059586826459\
> 0665546
>
> Puzzle: What are the next 6 decimal digits?

Solution: 141322

Method: Using "incgam", "eint1" and "moebius" this solution may
be found in less than 8 seconds, running at less than 3 GHz:

{F(n,k)=local(z=C*n^2);
incgam(1/2+k,z)/n^(2*k)+n*S[k]*eint(k,exp(-z),z);}

{eint(k,e,x)=if(k==1,eint1(x),(e-x*eint(k-1,e,x))/(k-1));}

{a(s)=sumdiv(s,d,if(d%2,moebius(d)*2^(s/d),0))/(2*s);}

{b(s)=if(s==1,0,a(s)-if(s%2,0,a(s/2)));}

{default(realprecision,550);P=31601;terms=60;gettime;
got=505;ext=6;C=Pi/4;S=sqrt(C)*vector(terms,k,C^k);
N=2*terms+1;Z=vector(N,s,if(s>1,zeta(s)*(1-1/2^s),1));
V=Vec(Pi/(4*cos(Pi*x/2+O(x^(N+2)))));L=vector(N,s,if(s%2,V[s],
suminf(n=0,F(4*n+1,s/2)-F(4*n+3,s/2))/gamma((s+1)/2)));t=0.5;
forprime(p=3,P,z=(-1)^((p-1)/2);t*=1-z/(p-1);r=1.;for(s=1,N,r/=p;
L[s]*=1-z*r;Z[s]*=1-(s>1)*r));for(s=1,N,t/=L[s]^a(s)*Z[s]^b(s));
t=floor(10^(got+ext)*t);print(t%(10^ext)" in " gettime" ms");}

141322 in 7507 ms

David
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