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Re: Sum of x/log(x) x=2..n

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  • Cino Hilliard
    Ok David, I give up. I searched and I searched and cannot find an interpretable (by me) routine I can convert to Pari or c/gmp. I can t afford Mathematica even
    Message 1 of 8 , Aug 7 9:42 AM
      Ok David, I give up.

      I searched and I searched and cannot find an interpretable (by me)
      routine I can convert to Pari or c/gmp.

      I can't afford Mathematica even at $295.

      How did you do it?

      I want to examine

      sm(m,n)=sum(x=m,n,x/log(x)-1/log(x)+a/log(x)^2-b/log(x)^3)
      for various values of m and n and hone in on a and b using
      known tables of Pi(x) and the approx R(x).

      a=1,b=1 m=2 gives pretty good approximations of Pi(n^2),
      about 1/2 the digits. Also it seems this is as good as
      Li(x).

      Are they asymptotic?

      It seems to me that the simple act of adding prime numbers
      and then noticing that the sum of these primes up to some
      number n is quite close to the count of the primes less than
      n^2 or Pi(n^2) should have been discovered hundreds of years
      ago.

      There should be a word for this. Often, knowledge replaces
      simple truths that lay hidden which much later are discovered
      with more knowledge.

      Basically, we always had a good approximation of Pi(x) by the
      simple operations of a sieve and addition. Somehow, the invention
      of natural logarithms provided the knowledge for a 15 year old (I
      doubt this and the add 1 to 100 story) to just start experimenting
      with a table of logs finding the log of a number and a count of
      primes from tables of primes. Had he not known logs, he may have
      experimented with the sum of the primes less than n in the tables
      of primes. After listing the sums, he then may have noticed that
      sum of prime < n ~ count of primes < n^2 which for large n,is a
      much better estimate of Pi(x) than x/(log(x)-1).

      For example,
      Pi(10^22) = 201467286689315906290
      SumP(sqrt(10^22)) = 201467077743744681014
      10^22/(log(10^22)-1) = 201381995844659893517.7646894


      In 1817 tables of primes grew to around less than 3 million. Too
      bad the table builders did not accumulate the primes as they built the tables.

      Nevertheless, my guess is a Savant could have accumulated the sums
      of the first 216,816 primes in a short time. Evan a couple of 8th
      grade classes could have done it in a few days. This is an example
      of an almost free lunch. A little addition early closely predicts
      a later truth.

      Per is summa nos victum,

      Cino

      --- In primenumbers@yahoogroups.com, "David Broadhurst" <d.broadhurst@...> wrote:
      >
      > > --- In primenumbers@yahoogroups.com,
      > > cino hilliard <hillcino368@> wrote:
      > > I am looking for a faster calculation for x/log(x).
      >
      > Euler-Maclaurin gave these results in less than a second:
      >
      > [ k, sum(x=2,10^k,x/log(x))]
      >
      > [ 9, 24739954333817884.765108796387854980062067356675901]
      > [10, 2220819603000810723.0256382787068599092733377762963]
      > [11, 201467286693222327323.50889662513497938846705904284]
      > [12, 18435599767384443378555.311612011169981547316897585]
      > [13, 1699246750872760069344915.8741205160589367732107901]
      > [14, 157589269275976389210121053.30825414331084987218682]
      > [15, 14692398897720462115561817776.152711329547881311518]
      > [16, 1376110866993766140634547918278.2822391520225219083]
      > [17, 129408626505158943971865401935043.41415238141439346]
      > [18, 12212914297619365466978260828820202.579301156342592]
      > [19, 1156251261026516898232106385426821880.5658457303344]
      > [20, 109778913489828302829174401609875888501.15351452821]
      > [21, 10449550362264587535492180923467208556377.216140121]
      > [22, 996973504768769817629382023798067858752836.80045752]
      >
      > 615 milliseconds
      >
      > David
      >
    • David Broadhurst
      ... http://physics.open.ac.uk/~dbroadhu/cert/eulmac.gp David
      Message 2 of 8 , Aug 7 10:47 AM
        --- In primenumbers@yahoogroups.com,
        "Cino Hilliard" <hillcino368@...> wrote:

        > Ok David, I give up.
        >
        > How did you do it?

        http://physics.open.ac.uk/~dbroadhu/cert/eulmac.gp

        David
      • David Broadhurst
        ... f(x,lx) = x/lx - 1/lx + a/lx^2 - b/lx^3; modify at will a=1; b=1; m=2; chosen values p50 terms=10;trunc=10^4;g=f(x,lx);d=[];bv=bernvec(terms+1);
        Message 3 of 8 , Aug 8 12:19 AM
          --- In primenumbers@yahoogroups.com,
          "Cino Hilliard" <hillcino368@...> wrote:

          > I want to examine
          > sm(m,n)=sum(x=m,n,x/log(x)-1/log(x)+a/log(x)^2-b/log(x)^3)
          > a=1,b=1,m=2

          f(x,lx) = x/lx - 1/lx + a/lx^2 - b/lx^3; \\ modify at will
          a=1; b=1; m=2; \\ chosen values

          \p50
          terms=10;trunc=10^4;g=f(x,lx);d=[];bv=bernvec(terms+1);

          {for(n=1,2*terms-1, \\ store odd derivatives
          g=deriv(g,x)+1/x*deriv(g,lx);if(n%2,d=concat(d,g)));}

          fx(x)=f(x,log(x));
          dx(y)=subst(subst(d,x,y),lx,log(y));

          {emac(m,n)=local(dm,dn);
          dm=dx(m);dn=dx(n);intnum(x=m,n,fx(x))+(fx(n)+fx(m))/2
          +sum(k=1,terms,bv[k+1]/((2*k)!)*(dn[k]-dm[k]));}

          s=sum(k=2,trunc-1,fx(k));

          {sm(m,n) = \\ as requested
          if(n>trunc&&trunc>m&&m>1,s-sum(k=2,m-1,fx(k))+emac(trunc,n));}

          for(k=10,22,print([k,sm(m,sqrtint(10^k))]))

          realprecision = 57 significant digits (50 digits displayed)
          [10, 455051173.78013348859945302791104662093667330309988]
          [11, 4118034570.4638268343263084222227560024582402009128]
          [12, 37607913583.517014991610186636613681414323919745603]
          [13, 346065399465.03248684994266161913298076743032790216]
          [14, 3204941752475.7266213978929589359007763992788467339]
          [15, 29844569450368.625194208305114011697818820369920647]
          [16, 279238341514801.87986838501363369093160129071417051]
          [17, 2623557157209703.2566630532132281009952796686474077]
          [18, 24739954285430064.167128134429584525396645626800130]
          [19, 234057667279245072.15948934429135642246573100181670]
          [20, 2220819602565566257.9962162533153260118661266449421]
          [21, 21127269485065195786.472528704950924409440892097567]
          [22, 201467286689267167012.50527750985870920146256019519]

          Comment: This is an inefficient method of fiddling with Li(x).
          Cino is trying to avoid integration. Yet it is needed for
          Euler-Maclaurin summation.

          David
        • Cino Hilliard
          Hi David, You did a lot of work here to develop emac(m,n).I appreciate it as it saved me a lot of time and possibly money. Thank you. ... Maybe. Anyway I don t
          Message 4 of 8 , Aug 8 11:13 PM
            Hi David,
            You did a lot of work here to develop emac(m,n).I appreciate
            it as it saved me a lot of time and possibly money.
            Thank you.

            David Broadhurst wrote:
            > Comment: This is an inefficient method of fiddling with Li(x).

            Maybe. Anyway I don't mind inefficiencies as I have to use them
            everyday like the QWERTY keyboard and all the right handed stuff
            around for this lefty.

            Surprizingly, the most efficient combination in the real world for moving a mass from point A to point B with the least expenditure
            of energy, is a human riding a bicycle on a flat surface. Then
            there is the hummingbird which is one of the the least efficient systems that has been around for a while.

            Nevertheless, Your emac(m,n) gets good mileage and is more
            accurate than Li(x) =-eint1(log(1/x)) and
            li(x,n) = lg=log(x);x*sum(k=1,n,(k-1)!/lg^k)

            gp > p23=1925320391606818006727
            gp > em23=emac(2,sqrt(10^23))
            %116 = 1925320391608063705941.14436225628
            gp > Li23=Li(10^23)
            %117 = 1925320391614054155138.78012956636
            gp > R23=R(10^23)
            %118 = 1925320391607837268776.09905063742
            gp > 1-em23/p23
            %122 = -6.4700878855011600657703221947361 E-13
            gp > 1-Li23/p23
            %123 = -3.7584125963269129092185767109212 E-12
            gp > 1-R23/p23
            %124 = -5.2939866712178918250597351383218 E-13

            For 10^22, emac is better than R(x).

            The actual sum of the primes < sqrt(n) ~ Pi(n) is not as
            accurate as emac but it is quite good wit RE below.

            For sum of primes < 10^n, See the b-file at
            http://www.research.att.com/~njas/sequences/A046731

            sump11 = 201467077743744681014
            gp > p22 = 201467286689315906290
            %137 = 201467286689315906290
            gp > 1. - sump11/p22
            %139 = 0.00000103711..
            This is magnitudes better than RE of 10^22/log(10^22)-1) =
            0.000423348356239

            I find this sum of primes <= sqrt(n) ~ primes <= n to be an
            amazing property of prime numbers and the integers.

            Bottomline, thanks to David, we have an eulermac function in
            Pari.

            Cheers and Roebuck,
            Cino
          • David Broadhurst
            ... Glad to help. ... Not so amazing when you realize that the integrals I1(N) = intnum(x=2, sqrt(N), x/log(x)) I2(N) = intnum(y=2, N , 1/log(y)) differ
            Message 5 of 8 , Aug 9 2:34 AM
              --- In primenumbers@yahoogroups.com,
              "Cino Hilliard" <hillcino368@...> wrote:

              > Hi David,
              > You did a lot of work here to develop emac(m,n).
              > I appreciate it as it saved me a lot of time and
              > possibly money. Thank you.

              Glad to help.

              > I find this sum of primes <= sqrt(n) ~ primes <= n to be
              > an amazing property of prime numbers and the integers.

              Not so amazing when you realize that the integrals

              I1(N) = intnum(x=2, sqrt(N), x/log(x))
              I2(N) = intnum(y=2, N , 1/log(y))

              differ only by a constant, no matter what the value of N.

              Proof: Transform the latter by setting y = x^2. Then
              I2(N) - I1(N) = - I1(2) = 1.92242131492155809316615998937951547...

              David
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