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Re: Conjecture Ludovicus V

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  • Robert Gerbicz
    This won t be log(2). Let q 2 prime (where q!=p) then x^2+(p-x)^2 != 0 mod q has got (q-2) solutions if 4|q-1 and q solutions if 4|q-3. Using this and the
    Message 1 of 6 , Nov 11, 2012
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      This won't be log(2). Let q>2 prime (where q!=p) then x^2+(p-x)^2 != 0 mod
      q has got (q-2) solutions if 4|q-1 and q solutions if 4|q-3. Using this and
      the prime number theorem my guess for the R value is
      0.5*prodprime(p=3,inf,1+(-1)^((p+1)/2)/(p-1))

      With PARI:
      r=0.5;forprime(p=3,4*10^9,r*=1+(-1)^((p+1)/2)/(p-1));r
      0.6864063439916482548130029679
      here probably the first five digits are correct, so R~0.68640


      [Non-text portions of this message have been removed]
    • djbroadhurst
      ... Nice work, Robert. http://tech.groups.yahoo.com/group/primenumbers/message/24612
      Message 2 of 6 , Nov 11, 2012
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        --- 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...

        David (corrected yet again)
      • djbroadhurst
        ... Here are more than 500 good digits of the relevant Hardy-Littlewood contant 0.686406731409123004556096348363509434089166550627
        Message 3 of 6 , Nov 12, 2012
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          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > 0.68640673...

          Here are more than 500 good digits of the relevant
          Hardy-Littlewood contant

          0.686406731409123004556096348363509434089166550627\
          87977896811707366392111335868511586385990346954399\
          18910971210114370836967066905693197374076733762228\
          99676181590249072148452433271956678446776133507821\
          66911238709121761546057993037867560421525052339604\
          30362058008970065288339284308924109839850906244857\
          05610181784829261874808914954658125072564596078857\
          97178488488284913596339147096072562959833915384281\
          95797768815771346844648575433020873782746429273861\
          06852378388023207827786598975144050262059586826459\
          0665546

          obtained in about 3 GHz-minutes.

          Puzzle: What are the next 6 decimal digits?

          David
        • mikeoakes2
          ... Regarding that nice 1960 paper: Daniel Shanks describes his eratosthenese-type sieve for factoring numbers of the form n^2+1 here:
          Message 4 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 5 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 6 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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