Loading ...
Sorry, an error occurred while loading the content.

What if Riemann's prime-counting formula was not the best?

Expand Messages
  • chrisdecorte
    What if another, easier to use formula, would fit better with pi(x)? Check it out at: http://www.slideshare.net/ChrisDeCorte1/better-prime-counting-formula
    Message 1 of 16 , Jul 28, 2013
    • 0 Attachment
      What if another, easier to use formula, would fit better with pi(x)?

      Check it out at:

      http://www.slideshare.net/ChrisDeCorte1/better-prime-counting-formula
    • djbroadhurst
      ... Perhaps it might be a good idea to find out what Riemann s formula is? Then try comparing your Excel power-law fits with Gram s convenient version of
      Message 2 of 16 , Jul 28, 2013
      • 0 Attachment
        --- In primenumbers@yahoogroups.com,
        "chrisdecorte" <chrisdecorte@...> wrote:

        > What if Riemann's prime-counting formula was not the best?

        Perhaps it might be a good idea to find out what Riemann's
        formula is? Then try comparing your Excel power-law fits
        with Gram's convenient version of Riemann's formula, which
        you will find in Eq (12) at

        http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html

        Here is R(10^k) for k = 1 to 10:

        R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
        for(k=1,10,print([k,round(R(10^k))]))
        [1, 5]
        [2, 26]
        [3, 168]
        [4, 1227]
        [5, 9587]
        [6, 78527]
        [7, 664667]
        [8, 5761552]
        [9, 50847455]
        [10, 455050683]

        Over to Excel :-)

        David
      • Chris De Corte
        Please find attached the comparison. To be more correct, I should also know the pi(x) so that I can fine-tune my formula as well.
        Message 3 of 16 , Jul 28, 2013
        • 0 Attachment
          Please find attached the comparison.
          To be more correct, I should also know the pi(x) so that I can fine-tune my formula as well.



          ________________________________
          From: djbroadhurst <d.broadhurst@...>
          To: primenumbers@yahoogroups.com
          Sent: Sunday, July 28, 2013 12:46 PM
          Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?




          --- In primenumbers@yahoogroups.com,
          "chrisdecorte" <chrisdecorte@...> wrote:

          > What if Riemann's prime-counting formula was not the best?

          Perhaps it might be a good idea to find out what Riemann's
          formula is? Then try comparing your Excel power-law fits
          with Gram's convenient version of Riemann's formula, which
          you will find in Eq (12) at

          http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html

          Here is R(10^k) for k = 1 to 10:

          R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
          for(k=1,10,print([k,round(R(10^k))]))
          [1, 5]
          [2, 26]
          [3, 168]
          [4, 1227]
          [5, 9587]
          [6, 78527]
          [7, 664667]
          [8, 5761552]
          [9, 50847455]
          [10, 455050683]

          Over to Excel :-)

          David




          ------------------------------------

          Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
          The Prime Pages : http://primes.utm.edu/

          Yahoo! Groups Links



          [Non-text portions of this message have been removed]
        • djbroadhurst
          ... http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html PS: Eric s Eq (11) is the last equation of Riemann s MS of 1859:
          Message 4 of 16 , Jul 28, 2013
          • 0 Attachment
            --- In primenumbers@yahoogroups.com,
            "djbroadhurst" <d.broadhurst@...> wrote:

            > > What if Riemann's prime-counting formula was not the best?
            >
            > Perhaps it might be a good idea to find out what Riemann's
            > formula is? Then try comparing your Excel power-law fits
            > with Gram's convenient version of Riemann's formula, which
            > you will find in Eq (12) at

            http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html

            PS: Eric's Eq (11) is the last equation of Riemann's MS of 1859:

            http://www.claymath.org/millennium/Riemann_Hypothesis/1859_manuscript/riemann1859.pdf

            David
          • Alan Powell
            Chris In the Wikipedia article titled ôPrime-counting functionö http://en.wikipedia.org/wiki/Prime-counting_function the value of Pi(10^24) is exactly
            Message 5 of 16 , Jul 28, 2013
            • 0 Attachment
              Chris

              In the Wikipedia article titled �Prime-counting function�

              http://en.wikipedia.org/wiki/Prime-counting_function

              the value of Pi(10^24) is exactly 18,435,599,767,349,200,867,866.

              Your �preferred formula�

              Pi(x) = alpha*x^beta with alpha=0.2083666 and beta=0.9294465 gives

              only 4,222,251,563,919,881,535,488 which is out by a factor of 4+ !!

              You may want to study some elementary books on Number Theory to
              understand why Riemann�s function gives the best asymptotic value
              as the number of primes tends towards infinity.

              Regards

              Alan



              [Non-text portions of this message have been removed]
            • Jens Kruse Andersen
              ... Your pi(x) approximation without relying on different formulas for different small x ranges is c(x) = 0.2083666*x^0.9294465. The prime number theorem shows
              Message 6 of 16 , Jul 28, 2013
              • 0 Attachment
                chrisdecorte wrote:
                > What if another, easier to use formula, would fit better with pi(x)?
                >
                > Check it out at:
                >
                > http://www.slideshare.net/ChrisDeCorte1/better-prime-counting-formula

                Your pi(x) approximation without relying on different formulas for
                different small x ranges is c(x) = 0.2083666*x^0.9294465.

                The prime number theorem shows pi(x)/c(x) tends to infinite but
                your paper says:
                "We especially wanted to focus on the set of primes that were
                probably unknown in the time of Riemann."

                So let's make a comparison with exactly computed pi(x) values.
                x/log(x) and x/(log(x)-1) are the commonly used approximations
                not using li(x) or long formulas.

                // http://oeis.org/A006880 Number of primes < 10^n
                {A006880=[4,25,168,1229,9592,78498,664579,5761455,50847534,
                455052511,4118054813,37607912018,346065536839,3204941750802,
                29844570422669,279238341033925,2623557157654233,24739954287740860,
                234057667276344607,2220819602560918840,21127269486018731928,
                201467286689315906290,1925320391606803968923,
                18435599767349200867866,176846309399143769411680];}

                c(x)=0.2083666*x^0.9294465;
                \p 5
                print("x, pi(x)/c(x), pi(x)/(x/log(x)), pi(x)/(x/(log(x)-1))");
                for(n=1,#A006880,x=10^n;p=A006880[n];\
                print("10^"n" "p/c(x)" "p/(x/log(x))" "p/(x/(log(x)-1))));

                x, pi(x)/c(x), pi(x)/(x/log(x)), pi(x)/(x/(log(x)-1))
                10^1 2.2583 0.92103 0.52103
                10^2 1.6604 1.1513 0.90129
                10^3 1.3126 1.1605 0.99250
                10^4 1.1296 1.1320 1.0091
                10^5 1.0372 1.1043 1.0084
                10^6 0.99851 1.0845 1.0060
                10^7 0.99447 1.0712 1.0047
                10^8 1.0142 1.0613 1.0037
                10^9 1.0530 1.0537 1.0029
                10^10 1.1086 1.0478 1.0023
                10^11 1.1802 1.0430 1.0019
                10^12 1.2679 1.0391 1.0015
                10^13 1.3725 1.0359 1.0013
                10^14 1.4953 1.0332 1.0011
                10^15 1.6381 1.0308 1.0010
                10^16 1.8030 1.0288 1.0008
                10^17 1.9928 1.0270 1.0007
                10^18 2.2107 1.0254 1.0006
                10^19 2.4604 1.0240 1.0006
                10^20 2.7463 1.0227 1.0005
                10^21 3.0735 1.0216 1.0005
                10^22 3.4479 1.0206 1.0004
                10^23 3.8762 1.0196 1.0004
                10^24 4.3663 1.0188 1.0004
                10^25 4.9273 1.0180 1.0003

                With 7 decimals you get enough wiggle room to come close in a small
                chosen range, but I'm not impressed by c(x) over a longer range.

                --
                Jens Kruse Andersen
              • Alan Powell
                Chris Yes if you use Excel to compute the 18,435,599,767,349,200,867,866 pairs of values (alpha, beta) your formula Pi(x) = alpha*x^beta will give accurate
                Message 7 of 16 , Jul 28, 2013
                • 0 Attachment
                  Chris

                  Yes if you use Excel to compute the 18,435,599,767,349,200,867,866
                  pairs of values (alpha, beta) your formula

                  Pi(x) = alpha*x^beta

                  will give accurate results up to x=10^24. The value of this
                  exercise escapes me.

                  Regards

                  Alan

                  From: Chris De Corte
                  Sent: Sunday, July 28, 2013 7:30 AM
                  To: djbroadhurst ; primenumbers@yahoogroups.com
                  Subject: Re: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?

                  Please find attached the comparison.
                  To be more correct, I should also know the pi(x) so that I can fine-tune my formula as well.

                  ________________________________
                  From: djbroadhurst <mailto:d.broadhurst%40open.ac.uk>
                  To: mailto:primenumbers%40yahoogroups.com
                  Sent: Sunday, July 28, 2013 12:46 PM
                  Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?


                  --- In mailto:primenumbers%40yahoogroups.com,
                  "chrisdecorte" <chrisdecorte@...> wrote:

                  > What if Riemann's prime-counting formula was not the best?

                  Perhaps it might be a good idea to find out what Riemann's
                  formula is? Then try comparing your Excel power-law fits
                  with Gram's convenient version of Riemann's formula, which
                  you will find in Eq (12) at

                  http://mathworld.wolfram.com/RiemannPrimeCountingFunction.html

                  Here is R(10^k) for k = 1 to 10:

                  R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
                  for(k=1,10,print([k,round(R(10^k))]))
                  [1, 5]
                  [2, 26]
                  [3, 168]
                  [4, 1227]
                  [5, 9587]
                  [6, 78527]
                  [7, 664667]
                  [8, 5761552]
                  [9, 50847455]
                  [10, 455050683]

                  Over to Excel :-)

                  David

                  ------------------------------------

                  Unsubscribe by an email to: mailto:primenumbers-unsubscribe%40yahoogroups.com
                  The Prime Pages : http://primes.utm.edu/

                  Yahoo! Groups Links

                  [Non-text portions of this message have been removed]





                  [Non-text portions of this message have been removed]
                • djbroadhurst
                  ... I think that Chris considers log to be a complicated function. Here is a perfect fit to Jens data, without using the dreaded log :
                  Message 8 of 16 , Jul 28, 2013
                  • 0 Attachment
                    --- In primenumbers@yahoogroups.com,
                    "Jens Kruse Andersen" <jens.k.a@...> wrote:

                    > So let's make a comparison with exactly computed pi(x) values.
                    > x/log(x) and x/(log(x)-1) are the commonly used approximations
                    > not using li(x) or long formulas.

                    I think that Chris considers "log" to be a complicated function.
                    Here is a perfect fit to Jens' data, without using the dreaded "log":

                    {data=[4,25,168,1229,9592,78498,664579,5761455,50847534,455052511,
                    4118054813,37607912018,346065536839,3204941750802,29844570422669,
                    279238341033925,2623557157654233,24739954287740860,234057667276344607,
                    2220819602560918840,21127269486018731928,201467286689315906290,
                    1925320391606803968923,18435599767349200867866,176846309399143769411680];}

                    Poly=polinterpolate(vector(25,k,10^k),data);
                    fit(n)=subst(Poly,x,n);

                    for(k=1,25,print1(fit(10^k)-data[k]","));
                    0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

                    Looks good, to log-haters? Ought to give a good
                    prediction to pi(10^26), surely :-?

                    print(fit(10.^26));
                    -6.2925202314581568492313228001263868142 E300

                    Hmm.. Maybe that pesky log function has it uses?

                    David
                  • WarrenS
                    Riemann s (& variant form) prime-counting formulas are the best in the sense they yield exact equality. Everywhere. Given various exact formulas and
                    Message 9 of 16 , Jul 28, 2013
                    • 0 Attachment
                      Riemann's (& variant form) prime-counting formulas are "the best" in the sense they
                      yield exact equality. Everywhere.

                      Given various exact formulas and algorithms for computing pi(x),
                      the "best" would seem to be the one permitting the fastest computation (maybe memory
                      consumption should also be taken into account).
                      And in fact, the fastest currently known algorithm (asymptotically) for computing
                      pi(x) is based on a form of Riemann, not combinatorial counting methods and not just generating primes. Also, for inexact computation of pi(x) to specified accuracy, this method still seems best known.

                      However, there is no reason to believe that further speedups are necessarily impossible
                      with further algorithmic/analytic ideas nobody has thought of yet.
                      It would help if anybody had the faintest idea how to prove lower bounds on the
                      computational complexity of counting primes.

                      I could continue. With Green-Tao, we know there are arbitrarily long arithmetic progressions with all-prime entries. How computationally complex is it to find an N-term example? Well... it seems to be extremely hard... quite likely humanity will
                      never find an example with N=50... but I am unaware of any
                      reasonable lower bounds, or upper bounds, on this computational complexity!
                    • Chris De Corte
                      Alan, I said in my document that I only made a calculation up to 49978001 because I don t have other data available. If you all agree on a new trial then
                      Message 10 of 16 , Jul 28, 2013
                      • 0 Attachment
                        Alan,

                        I said in my document that I only made a calculation up to 49978001 because I don't have other data available.

                        If you all agree on a new trial then please provide me with 10 N, 10 pi(x) and  10 best Riemann's approximations and I will try to calculate a new alpha and beta.

                        Best,
                        Chris


                        ________________________________
                        From: Alan Powell <AlanPowell@...>
                        To: primenumbers@yahoogroups.com
                        Cc: Chris De Corte <chrisdecorte@...>
                        Sent: Sunday, July 28, 2013 1:42 PM
                        Subject: Re: What if Riemann's prime-counting formula was not the best?



                        Chris
                         
                        In the Wikipedia article titled “Prime-counting function”
                         
                          http://en.wikipedia.org/wiki/Prime-counting_function%c2%a0
                         
                        the value of Pi(10^24) is exactly 18,435,599,767,349,200,867,866.
                         
                        Your “preferred formula”
                         
                          Pi(x) = alpha*x^beta  with alpha=0.2083666 and beta=0.9294465
                        gives
                         
                        only 4,222,251,563,919,881,535,488  which is out by a
                        factor of 4+ !!
                         
                          You may want to study some elementary books on Number Theory to
                        understand why Riemann’s function gives the best asymptotic value
                        as the number of primes tends towards infinity.
                         
                        Regards
                         
                        Alan

                        [Non-text portions of this message have been removed]
                      • djbroadhurst
                        ... Please use plain text in messages to this list, else you may not be understood. Here I provide [N, pi(10^N), R(10^N)] [10, 455052511, 455050683] [11,
                        Message 11 of 16 , Jul 28, 2013
                        • 0 Attachment
                          --- In primenumbers@yahoogroups.com,
                          Chris De Corte <chrisdecorte@...> wrote:

                          > please provide me with 10 N, 10 pi(x) and  10 best Riemann's
                          > approximations and I will try to calculate a new alpha and beta.

                          Please use plain text in messages to this list, else
                          you may not be understood. Here I provide

                          [N, pi(10^N), R(10^N)]
                          [10, 455052511, 455050683]
                          [11, 4118054813, 4118052495]
                          [12, 37607912018, 37607910542]
                          [13, 346065536839, 346065531066]
                          [14, 3204941750802, 3204941731602]
                          [15, 29844570422669, 29844570495887]
                          [16, 279238341033925, 279238341360977]
                          [17, 2623557157654233, 2623557157055978]
                          [18, 24739954287740860, 24739954284239494]
                          [19, 234057667276344607, 234057667300228940]
                          [20, 2220819602560918840, 2220819602556027015]
                          [21, 21127269486018731928, 21127269485932299724]
                          [22, 201467286689315906290, 201467286689188773625]
                          [23, 1925320391606803968923, 1925320391607837268776]
                          [24, 18435599767349200867866, 18435599767347541878147]
                          [25, 176846309399143769411680, 176846309399141934626966]

                          where

                          R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

                          David
                        • Chris De Corte
                          Hi, This is the result: N Pi(x) R(x) Power 10 1E+10 455052511 455050683 420213980 11 1E+11 4118054813 4118052495 3951369042 12 1E+12 37607912018 37607910542
                          Message 12 of 16 , Jul 28, 2013
                          • 0 Attachment
                            Hi,

                            This is the result:

                            N Pi(x) R(x) Power
                            10 1E+10 455052511 455050683 420213980
                            11 1E+11 4118054813 4118052495 3951369042
                            12 1E+12 37607912018 37607910542 37155635094
                            13 1E+13 3.46066E+11 3.46066E+11 349383012475
                            14 1E+14 3.20494E+12 3.20494E+12 3285329105477
                            15 1E+15 2.98446E+13 2.98446E+13 30892707847616
                            16 1E+16 2.79238E+14 2.79238E+14 290491262067780
                            17 1E+17 2.62356E+15 2.62356E+15 2731556383920000
                            18 1E+18 2.474E+16 2.474E+16 25685455133563200
                            19 1E+19 2.34058E+17 2.34058E+17 241526262940073000
                            20 1E+20 2.22082E+18 2.22082E+18 2271127195778980000
                            21 1E+21 2.11273E+19 2.11273E+19 21355933208335000000
                            22 1E+22 2.01467E+20 2.01467E+20 200814769003914000000
                            23 1E+23 1.92532E+21 1.92532E+21 1888307621900430000000
                            24 1E+24 1.84356E+22 1.84356E+22 17756192398666400000000
                            25 1E+25 1.76846E+23 1.76846E+23 166965575334147000000000

                            alpha 0.07774984570
                            beta 0.9732770961
                            correlation R(x) 1.00000000000
                            correlation power 0.999997766
                            So, yes, Riemann function is better in this test.
                            I also attach the excel for those who can open it.

                            I based my testing on a book I was reading about unsolved problems that wrote that Riemann had used the formula x/ln(x) which was obviously an oversimplification.

                            I am sorry if I wasted your guys day.

                            Thanks & good night,
                            Chris



                            ________________________________
                            From: djbroadhurst <d.broadhurst@...>
                            To: primenumbers@yahoogroups.com
                            Sent: Sunday, July 28, 2013 9:11 PM
                            Subject: [PrimeNumbers] Re: What if Riemann's prime-counting formula was not the best?




                            --- In primenumbers@yahoogroups.com,
                            Chris De Corte <chrisdecorte@...> wrote:

                            > please provide me with 10 N, 10 pi(x) and  10 best Riemann's
                            > approximations and I will try to calculate a new alpha and beta.

                            Please use plain text in messages to this list, else
                            you may not be understood. Here I provide

                            [N, pi(10^N), R(10^N)]
                            [10, 455052511, 455050683]
                            [11, 4118054813, 4118052495]
                            [12, 37607912018, 37607910542]
                            [13, 346065536839, 346065531066]
                            [14, 3204941750802, 3204941731602]
                            [15, 29844570422669, 29844570495887]
                            [16, 279238341033925, 279238341360977]
                            [17, 2623557157654233, 2623557157055978]
                            [18, 24739954287740860, 24739954284239494]
                            [19, 234057667276344607, 234057667300228940]
                            [20, 2220819602560918840, 2220819602556027015]
                            [21, 21127269486018731928, 21127269485932299724]
                            [22, 201467286689315906290, 201467286689188773625]
                            [23, 1925320391606803968923, 1925320391607837268776]
                            [24, 18435599767349200867866, 18435599767347541878147]
                            [25, 176846309399143769411680, 176846309399141934626966]

                            where

                            R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

                            David



                            ------------------------------------

                            Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
                            The Prime Pages : http://primes.utm.edu/

                            Yahoo! Groups Links



                            [Non-text portions of this message have been removed]
                          • djbroadhurst
                            ... That was, of course, a log-lover s spoof on log-haters. However, this log-hater: http://arxiv.org/pdf/1307.4444.pdf attempts to fix up the obvious lunacy
                            Message 13 of 16 , Jul 28, 2013
                            • 0 Attachment
                              --- In primenumbers@yahoogroups.com,
                              "djbroadhurst" <d.broadhurst@...> wrote:

                              > Poly=polinterpolate(vector(25,k,10^k),data);

                              That was, of course, a log-lover's spoof on log-haters.

                              However, this log-hater:
                              http://arxiv.org/pdf/1307.4444.pdf
                              attempts to fix up the obvious lunacy of polynomial
                              approximation of pi(x).

                              Andrey Kulsha may offer us tighter conjectural
                              bounds on pi(10^26) ?

                              David
                            • djbroadhurst
                              ... On the assumption that Delta(x) defined by Andrey in http://www.primefan.ru/stuff/primes/table.html#theory continues to satisfy abs(Delta(x))
                              Message 14 of 16 , Jul 28, 2013
                              • 0 Attachment
                                --- In primenumbers@yahoogroups.com,
                                "djbroadhurst" <d.broadhurst@...> wrote:

                                > Andrey Kulsha may offer us tighter conjectural
                                > bounds on pi(10^26) ?

                                On the assumption that Delta(x) defined by Andrey in
                                http://www.primefan.ru/stuff/primes/table.html#theory
                                continues to satisfy abs(Delta(x)) < 1, I estimate that

                                pi(10^26) = 1699246750872419991992147 +/- 167036339194

                                David (subject to error, as ever)
                              • Chroma
                                djbroadhurst wrote: ... For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms, Much faster can be calculated
                                Message 15 of 16 , Jul 30, 2013
                                • 0 Attachment
                                  djbroadhurst wrote:>
                                  >
                                  > [N, pi(10^N), R(10^N)]
                                  ........
                                  > [25, 176846309399143769411680, 176846309399141934626966]
                                  >
                                  > where
                                  >
                                  > R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
                                  >
                                  For large values ​​of x, this algorithm is inconvenient, eg for x = 10^250 requires over 1868 terms,
                                  Much faster can be calculated as

                                  pi(x) ~= pli(x) = round(Li(x) - 1/2 Li(sqrt(x)))

                                  where Li(x) is the Logarithm integral

                                  pli(10^25) = 176846309399141938590795
                                  (pli(10^25)/R(10^25)) - 1 = 2 10^-17
                                  (pli(10^25)/pi(10^25)) -1 = -1 10^-14

                                  pli{10^250)= 1740206254656916846774941665048386410178028975968929264655269395003484\
                                  7365084787720410883002915274182213664956284195372937010842285191263145\
                                  7678993892420170619475710388189158537825404886895382231933346054713467\
                                  85875358018952542776800464839768387582

                                  --
                                  marian otremba
                                • djbroadhurst
                                  ... No. The Gram formula is still very convenient at this size. Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:
                                  Message 16 of 16 , Jul 30, 2013
                                  • 0 Attachment
                                    --- In primenumbers@yahoogroups.com,
                                    Chroma <chromatella@...> wrote:

                                    >> R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));
                                    > For large values of x, this algorithm is inconvenient,
                                    > eg for x = 10^250 requires over 1868 terms

                                    No. The Gram formula is still very convenient at this size.
                                    Pari-GP, gives the exact value of R(10^250) in 0.1 seconds:

                                    R(x)=round(1+suminf(k=1,log(x)^k/(zeta(k+1)*k*k!)));

                                    {default(realprecision,260);print(R(10^250));
                                    print(" took "gettime" milliseconds");

                                    17402062546569168467749416650483864101780289759689292646552693950034847365084787720410883002915274182213664956284195372937010842285191263145767899389242017061947571038442681072462756632213511422607548574658029047365218974809766827365028215685475746
                                    took 98 milliseconds

                                    Perhaps you are paying for inferior software?
                                    If so, the general rule is: the less you pay,
                                    the better the deal.

                                    Pari-GP is totally free and hence rather hard to beat :-)

                                    David
                                  Your message has been successfully submitted and would be delivered to recipients shortly.