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Re: Conjecture Ludovicus V

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  • mikeoakes2
    ... Regarding that nice 1960 paper: Daniel Shanks describes his eratosthenese-type sieve for factoring numbers of the form n^2+1 here:
    Message 1 of 6 , Nov 12, 2012
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com,
      > Robert Gerbicz <robert.gerbicz@> wrote:
      >
      > > 0.5*prodprime(p=3,inf,1+(-1)^((p+1)/2)/(p-1))
      >
      > Nice work, Robert.
      >
      > http://tech.groups.yahoo.com/group/primenumbers/message/24612
      > http://www.ams.org/journals/mcom/1960-14-072/S0025-5718-1960-0120203-6/S0025-5718-1960-0120203-6.pdf
      >
      > 0.68640673...

      Regarding that nice 1960 paper: Daniel Shanks describes his eratosthenese-type sieve for factoring numbers of the form n^2+1 here:
      http://www.ams.org/journals/mcom/1959-13-066/S0025-5718-1959-0105784-2/S0025-5718-1959-0105784-2.pdf

      Mike
    • djbroadhurst
      ... 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 2 of 6 , Nov 12, 2012
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        --- In primenumbers@yahoogroups.com,
        "mikeoakes2" <mikeoakes2@...> wrote:

        > http://www.ams.org/journals/mcom/1959-13-066/S0025-5718-1959-0105784-2/S0025-5718-1959-0105784-2.pdf

        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
      • djbroadhurst
        ... 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 3 of 6 , Nov 17, 2012
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          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > 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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