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Re: [PrimeNumbers] New file uploaded to primenumbers

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  • Phil Carmody
    It s nice to see such clearly presented ideas, definitely worth having a look. I ve noticed Marcel take a shot at one of the final theories, which implies that
    Message 1 of 18 , Feb 5, 2001
      It's nice to see such clearly presented ideas, definitely worth having a look. I've noticed Marcel take a shot at one of the final theories, which implies that there were no obvious bloopers in the first part.
      The early proofs are to an amateur like myself quite convincing - I particularly like the Twin prime theorem.

      However, one thing that this has higlighted is my weakness in absolute rigor, and I am a little wary about a couple of things.
      - firstly that just because something is true for every finite number does not (IIRC) mean that it is true as a limit.
      [ The example I remember from University being that of the 'staircase diagonal'. For _every_ granularity of staircase, the length was 2, but the limiting length was 1.414... ]
      - other things I can't quite formulate, just an unease about the 'rounding to integer' that takes place everywhere. I can't describe my worries yet, so I'll scratch a few things on paper first, and work out what I don't like. (However, I think that replacement of n by n# would make these worries go away.)


      I really would like to see another critical eye cast over these. In particular, it might well be that my worry about limits are either unfounded or irrelevant (i.e. the proofs can be reformulated without the concept of a limit).

      Phil

      On Sun, 04 February 2001, primenumbers@yahoogroups.com wrote:
      > Hello,
      >
      > This email message is a notification to let you know that
      > a file has been uploaded to the Files area of the primenumbers
      > group.
      >
      > File : /Josip Novakovic's Proof/primes.doc
      > Uploaded by : novakovic_josip@...

      Mathematics should not have to involve martyrdom;
      Support Eric Weisstein, see http://mathworld.wolfram.com
      Find the best deals on the web at AltaVista Shopping!
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    • Phil Carmody
      This theorem states that PrimePi(x) = -1 + Sum[k = 2..x] foo(k) Where foo(k) is a function which is 1 for primes or 4, and 0 otherwise. In particular: foo(k) =
      Message 2 of 18 , Aug 21, 2001
        This theorem states that
        PrimePi(x) = -1 + Sum[k = 2..x] foo(k)

        Where foo(k) is a function which is 1 for primes or 4, and 0 otherwise.

        In particular:
        foo(k) = bar(k, baz(k) )

        Where baz(k) is not in a closed form, and may require up to k steps to calculate.

        I.e. this is /quadratic/ time method, or O(N^2)

        Wheel factoring up to sqrt(N) is a O(sqrt(N)) method of testing primality. i.e. Prime counting can be done in O(N^1.5) by the most basic method known - by counting primes.

        What on earth does introducing contrived (Caran D'Ache! :-) ) functions gain us, in any way, in this regard?

        Or is it just trying to do things in the most complicated way possible - like my Obfuscated C Code?

        Phil





        Mathematics should not have to involve martyrdom;
        Support Eric Weisstein, see http://mathworld.wolfram.com
        Find the best deals on the web at AltaVista Shopping!
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      • Cletus Emmanuel
        Steven, Thank you for the upload of the Carol & Kynea Primes. Paul had mentioned that we upload a file to the primenumber group and I suggested that he look
        Message 3 of 18 , Mar 11, 2002
          Steven, Thank you for the upload of the Carol & Kynea Primes. Paul had
          mentioned that we upload a file to the primenumber group and I suggested
          that he look into it. Since you've done it then I guess that he will
          maintain it. One suggestion though, I think that the numbers should be
          ranked from highest to lowest and the rank be listed next to each number.
          Also, it is understood that I found the ones for K>15000, but we need to
          know the full names of the other two gentlemen as well. So I am suggesting
          that their full names be included. Thirdly, i think that the number of
          digits in the numbers should also be given. We can follow a format just
          like Chris Caldwell's Largest known primes list. We can also
          includeinformation like possible divisors of these numbers are == +/-1 mod
          8, just to name a few. We can even include a little biography of the
          discoverers...

          Let me know what you all think.
          Cletus



          >
          >Hello,
          >
          >This email message is a notification to let you know that
          >a file has been uploaded to the Files area of the primenumbers
          >group.
          >
          > File : /Prime Tables/Carol&Kynea.txt
          > Uploaded by : harvey563 <sharvey@...>
          > Description : primes of form 4^n+-2^n-1
          >
          >You can access this file at the URL
          >
          >http://groups.yahoo.com/group/primenumbers/files/Prime%20Tables/Carol%26Kynea.txt
          >
          >To learn more about file sharing for your group, please visit
          >
          >http://help.yahoo.com/help/us/groups/files
          >
          >Regards,
          >
          >harvey563 <sharvey@...>
          >
          >
          >
          >
          >


          _________________________________________________________________
          Send and receive Hotmail on your mobile device: http://mobile.msn.com
        • nonsemantic@aol.com
          Mr Brennan said: It looks like the Pollard rho algorithm, but it appears to have some minor changes from the canonical implementation. I didn t try running
          Message 4 of 18 , Sep 18 9:55 AM
            Mr Brennan said:

            "It looks like the Pollard rho algorithm, but it appears to have
            some minor changes from the "canonical" implementation.
            I didn't try running it -- I just had a quick glance at the code."

            The code was an attempt at a modifcation of the Number Field Sieve method (find w and v such that w^2 mod operand = v^2, then find the gcd of w - v and operand). It looks for a solution in the area where w is less than the integer square root of the operand plus the remainder. So, if we start with 21 as our operand, it'll begin at:

            4 + (21 - 16)
            4 + 5 = 9

            Now, 9^2 mod 21 = 81 mod 21 = (81 - 63) = 18

            the integer square root of 18 is 4, thus
            w - v = 9 - 4 = 5, and gcd of 5 and 21 is 1.

            The program, when it sees a gcd of 1 or 2 (2 I use to filter out any even number that may show up as a factor), will then loop through the same operation, only with a different w. The new
            value of w is determined by an algorithm that takes the old w and divides it by 10, then subtracts the result from w. Thus we
            have 9 / 10 = .9 In this instance, the program will go into a loop that iterates through all the values between 9 and 1. To give a flavor of what happens with a real-world number, here's the output of the program (with the value of w being reported each time through the loop) for a semi-random number:

            Now factoring 1234567894891...
            0
            1351681
            1216513
            1081345
            1216513
            1202997
            1189481
            1175965
            1162449
            1148933
            1135417
            1121901
            1108385
            1121901
            1120550
            1119199
            1117848
            1116497
            1115146
            1113795
            1112444
            1111093
            1112444
            1112309
            1112174
            1112039
            1111904
            1111769
            1111634
            1111499
            1111364
            1111229
            1111094
            1111229
            1111216
            1111203
            1111190
            1111177
            1111164
            1111151
            1111138
            1111125
            1111112
            1111099
            1111112
            1111111
            1111111
            1111110
            1111109
            1111108
            1111107
            1111106
            1111105
            1111104
            1111103
            1111102
            No answer found.

            If you look at the tenth entry from the end, you'll notice that it approaches the integer square root of the input number (the square root of 1234567894891 is 1111111.10825651). Anyway, for certain numbers (I'd really like to know if they have some common properties...) the program will find a factor of the input number operand in its descent towards the square root. I don't know exactly why that occurs, but I plan on investigating it further when I have more time (I'm suffering from Very Large Work Project syndrome that's cutting into my recreational programming and research time). Thanks for the info, btw, about Pollard Rho - I imagine that many of the answers I'm looking for are in the theory behind that algorithm.

            -Andrew Plewe-
          • David Broadhurst
            Andrew: The {m,n} numbers were presumably chosen using an isprime test. If one keeps {m,n} as close as possible, to make h=m^2-m*n+n^2 as small as possible,
            Message 5 of 18 , Oct 16, 2002
              Andrew: The {m,n} numbers were presumably chosen using
              an "isprime" test.

              If one keeps {m,n} as close as possible,
              to make h=m^2-m*n+n^2 as small as possible,
              for fixed d=m+n, then one gets:

              {forprime(d=3,10^4,t=0;ns=(d-1)/2;
              forstep(n=ns,1,-1,
              if(t==0,m=d-n;h=d^2-3*m*n;
              if(isprime(h),t=1;print([m,n,h,d]))));
              if(t==0,print("fail" d)))}

              [2, 1, 3, 3]
              [3, 2, 7, 5]
              [4, 3, 13, 7]
              [6, 5, 31, 11]
              [7, 6, 43, 13]
              [9, 8, 73, 17]
              [11, 8, 97, 19]
              [13, 10, 139, 23] [?]
              [15, 14, 211, 29]
              [16, 15, 241, 31]
              [20, 17, 349, 37]
              [21, 20, 421, 41]
              ....
              [no failures to up to 10^4]

              [?] Like you I am mystified why the author selected
              [16, 7, 193, 23] with {16,7}
              > presi in modo "distinto"
              instead of the closest pair {13,10}

              I guess the conjecture is that it can always
              be done one way or another.

              The heuristics are resonable:
              you get (d-1)/2 attempts at a prime h
              with d^2 > h > d^2/2.

              PNT says that this is "very easy on average",
              but that is quite different from proving
              what I infer to be:

              una congettura?

              David
            • David Broadhurst
              ... To be clear, let me make a conjecture which I believe to be highly probable, but cannot prove. Conjecture: For every integer x with gcd(x,6)=1 there exists
              Message 6 of 18 , Oct 16, 2002
                Edgar:

                > Conjecture is not one.
                > They are convinced that its development is a theorem.

                > Non =E8 una congettura. Sono convinto che il suo
                > sviluppo sia un teorema.

                To be clear, let me make a conjecture
                which I believe to be highly probable,
                but cannot prove.

                Conjecture: For every integer x with gcd(x,6)=1
                there exists an integer y such that
                h=(x^2+3*y^2)/4
                is a prime less than x^2.

                Comment: Set x=m+n, y=m-n to get your quadratic form
                h=(m^3+n^3)/(m+n)=m^2-m*n+n^2
                which I prefer to diagonalize.
                Note that I conjecture that there is at least one
                prime of your form when we fix m+n to be any
                odd number not divisible by 3.

                I cannot prove my conjecture.
                You can prove this thing?

                Non posso dimostrare la mia congettura.
                Potete dimostrare questa cosa?

                David
              • David Broadhurst
                Thanks Jim: All well now. I also checked that a signed cert goes OK: the trailing [Signature] 1$=0DFF68B645EEB5780475FFD833CD8E984E73D39D
                Message 7 of 18 , Nov 14, 2002
                  Thanks Jim: All well now. I also checked that a signed
                  cert goes OK: the trailing
                  [Signature]
                  1$=0DFF68B645EEB5780475FFD833CD8E984E73D39D
                  2$=58B83826080F6CF6CB62CCC14288EC15D4EB7FF1
                  was wisely ignored.
                  David
                • David Broadhurst <d.broadhurst@open.ac.u
                  ... http://groups.yahoo.com/group/primenumbers/files/Symbols.jpg 1) integral round closed contour in the complex plane 2) principal value of integral with a
                  Message 8 of 18 , Dec 14, 2002
                    Zaher:

                    > Description : symbols i want thier description
                    http://groups.yahoo.com/group/primenumbers/files/Symbols.jpg

                    1) integral round closed contour in the complex plane

                    2) principal value of integral with a pole in the integrand

                    3) [looks like an integral needing viagra, but is probably]
                    an integral avoiding a pole on the real axis
                    by a small semicircular detour to the upper half plane,
                    in which case its real part is as in (2) and
                    there is an imaginary part i*Pi*residue_at_pole
                    [assuming that f(z)*=f(z*)]

                    But none of these Cauchy symbols
                    has much to do with primes.

                    David
                  • mikeoakes2@aol.com
                    In a message dated 12/01/03 03:54:29 GMT Standard Time, ... Andrey: there would appear to be an error in your second formula for N_0(T) [ another exact form ],
                    Message 9 of 18 , Jan 12, 2003
                      In a message dated 12/01/03 03:54:29 GMT Standard Time,
                      primenumbers@yahoogroups.com writes:


                      > File : /Articles/zetazerosdistrib.pdf
                      > Uploaded by : andrey_601 <Andrey_601@...>
                      > Description : On the distribution of zeta zeros
                      >
                      > You can access this file at the URL
                      >
                      >
                      >
                      >
                      > http://groups.yahoo.com/group/primenumbers/files/Articles/zetazerosdistrib.pdf
                      >
                      >

                      Andrey: there would appear to be an error in your second formula for N_0(T)
                      ["another exact form"], as n is a free variable and undefined. Is there a
                      "sigma over n" missing, or what?

                      Mike Oakes


                      [Non-text portions of this message have been removed]
                    • Andrey Kulsha
                      ... The sigma is over prime powers p^n
                      Message 10 of 18 , Jan 12, 2003
                        > > http://groups.yahoo.com/group/primenumbers/files/Articles/zetazerosdistrib.pdf
                        > >
                        > >
                        >
                        > Andrey: there would appear to be an error in your second formula for N_0(T)
                        > ["another exact form"], as n is a free variable and undefined. Is there a
                        > "sigma over n" missing, or what?

                        The sigma is over prime powers p^n < x. Of course, n is positive integer, as defined in the first formula and can be seen from context.

                        Best wishes,

                        Andrey


                        [Non-text portions of this message have been removed]
                      • mikeoakes2@aol.com
                        In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@uow.edu.au ... Andrew: These are very interesting results! However, the phrase are almost
                        Message 11 of 18 , Jan 20, 2003
                          In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@...
                          writes:


                          > Anyway, at the top of the list are a trio of zeros with very low
                          > imaginary part, beating the 2.817...*I zero found a while back by
                          > David Broadhurst. These were initially found with 9 sig. figure
                          > precision, and then computed to greater precision with more terms of
                          > the series, so are almost certainly real.
                          >

                          Andrew: These are very interesting results!

                          However, the phrase "are almost certainly real" implies that you don't
                          believe the imag part is 0.002... but is rather 0.0, so to quote results to 9
                          figures is a bit misleading... To put it another way: what precision _are_
                          you claiming?

                          Could you maybe expand on what "more terms of the series" means: how many
                          terms did you take, and of what series? This would help others trying to
                          duplicate/refine your figures.

                          Mike Oakes



                          [Non-text portions of this message have been removed]
                        • andrew_j_walker <ajw01@uow.edu.au>
                          ... of ... don t ... results to 9 ... _are_ ... The bit in quotes means I believe they are almost certainly actual zeros, in the earlier messages there was
                          Message 12 of 18 , Jan 20, 2003
                            --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
                            > In a message dated 20/01/03 00:41:59 GMT Standard Time, ajw01@u...
                            > writes:
                            >
                            >
                            > > Anyway, at the top of the list are a trio of zeros with very low
                            > > imaginary part, beating the 2.817...*I zero found a while back by
                            > > David Broadhurst. These were initially found with 9 sig. figure
                            > > precision, and then computed to greater precision with more terms
                            of
                            > > the series, so are almost certainly real.
                            > >
                            >
                            > Andrew: These are very interesting results!
                            >
                            > However, the phrase "are almost certainly real" implies that you
                            don't
                            > believe the imag part is 0.002... but is rather 0.0, so to quote
                            results to 9
                            > figures is a bit misleading... To put it another way: what precision
                            _are_
                            > you claiming?

                            The bit in quotes means I believe they are almost certainly "actual"
                            zeros, in the earlier messages there was one case in which a zero
                            I found wasn't actual/real as pari's suminf doesn't like this series.
                            So although Newton's method converged, the value was bogus! Now I use
                            the sum function.
                            They don't have any nice Re(z)=0.5 properties like the normal zeta
                            function!
                            > Could you maybe expand on what "more terms of the series" means: how
                            many
                            > terms did you take, and of what series? This would help others
                            trying to
                            > duplicate/refine your figures.
                            >
                            The series I used is formula (4) in
                            http://mathworld.wolfram.com/PrimeZetaFunction.html
                            the normal (1) doesn't converge over a wide range.

                            I found for the initial zero finding, using /p9 in pari, 10000 terms
                            was enough. Looking at the sum of the series for terms 10001 to 15000
                            confirms this. Generally more terms are required for smaller Re(z),
                            and perhaps the number of terms needed also slowly increases with
                            Im(z), but it has been ok in the range I looked at. For Re(z)<0.005
                            the number of terms would be even greater.

                            You'll also see a number of the zeros have been computed to greater
                            precision ( /p28 or the default in pari). So far I've found 20000
                            terms has been enough for this. Comparing the more accurate results
                            with the /p9 value, I believe the zeros should be all accurate except
                            in the last digit of the real and imaginary parts. This should also be
                            the case for the /p28 values.

                            Andrew
                          • mikeoakes2@aol.com
                            In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@uow.edu.au ... Andrew: your table of zeros is fascinating. But I continue to be worried: all the
                            Message 13 of 18 , Jan 25, 2003
                              In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@...
                              writes:


                              > The series I used is formula (4) in
                              > http://mathworld.wolfram.com/PrimeZetaFunction.html
                              > the normal (1) doesn't converge over a wide range.
                              >
                              >

                              Andrew: your table of zeros is fascinating.
                              But I continue to be worried: all the references I've found that give the
                              formula
                              prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
                              quote the range of validity one would expect, namely: Re(n) > 1.

                              You are assuming this formula is valid for 0 < Re(n) < 1 also.

                              For Re(n) < 1, some of the zeta()'s in this sum have Re(argument) < 1, and
                              are therefore analytic continuations of the "original" zeta() (defined in
                              terms of the Euler product), and it is only the latter definition which
                              allows the Mobius inversion formula to be used.

                              For n = 1/m (m >= 2), one of the zeta()'s has argment 1.0, and so is infinite.

                              So, although you are computing a formal sum of zeta functions which converges
                              to prime_zeta(n) for Re(n) > 1, it isn't obvious that, for Re(n) < 1, said
                              sum is in fact prime_zeta(n).

                              I also have worries about the computation of ln(zeta(k*n)) when n is not
                              real: remembering that ln(z), for complex z, is multi-valued, with period
                              2*pi*i, are you sure the correct value is always being taken by the software
                              you use?

                              Mike



                              [Non-text portions of this message have been removed]
                            • David Broadhurst <d.broadhurst@open.ac.u
                              ... Yes, but I quickly found and Andrew massively confirmed that if you are prepared to live with conditional convergence and can smartly exploit Pari-Gp, then
                              Message 14 of 18 , Jan 25, 2003
                                Mike Oakes said to Andrew Walker:

                                > But I continue to be worried: all the references
                                > I've found that give the formula
                                > prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
                                > quote the range of validity one would expect,
                                > namely: Re(n) > 1.

                                Yes, but I quickly found and Andrew massively confirmed
                                that if you are prepared to live with conditional convergence
                                and can smartly exploit Pari-Gp, then you handle smaller Re(n),
                                with a price that increases as you get closer to zero.

                                What goes for you is the random process im moebius :-)

                                David
                              • andrew_j_walker <ajw01@uow.edu.au>
                                ... give the ...
                                Message 15 of 18 , Jan 25, 2003
                                  --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
                                  > In a message dated 21/01/03 01:01:02 GMT Standard Time, ajw01@u...
                                  > writes:
                                  >
                                  >
                                  > > The series I used is formula (4) in
                                  > > http://mathworld.wolfram.com/PrimeZetaFunction.html
                                  > > the normal (1) doesn't converge over a wide range.
                                  > >
                                  > >
                                  >
                                  > Andrew: your table of zeros is fascinating.
                                  > But I continue to be worried: all the references I've found that
                                  give the
                                  > formula
                                  > prime_zeta(n) = {sigma k=1 to oo}(mu(k)/k)*ln(zeta(k*n))
                                  > quote the range of validity one would expect, namely: Re(n) > 1.
                                  >
                                  > You are assuming this formula is valid for 0 < Re(n) < 1 also.
                                  >
                                  > For Re(n) < 1, some of the zeta()'s in this sum have Re(argument)
                                  < 1, and
                                  > are therefore analytic continuations of the "original" zeta()
                                  (defined in
                                  > terms of the Euler product), and it is only the latter definition
                                  which
                                  > allows the Mobius inversion formula to be used.
                                  >
                                  > For n = 1/m (m >= 2), one of the zeta()'s has argment 1.0, and so
                                  is infinite.
                                  >
                                  > So, although you are computing a formal sum of zeta functions
                                  which converges
                                  > to prime_zeta(n) for Re(n) > 1, it isn't obvious that, for Re(n) <
                                  1, said
                                  > sum is in fact prime_zeta(n).
                                  >
                                  > I also have worries about the computation of ln(zeta(k*n)) when n
                                  is not
                                  > real: remembering that ln(z), for complex z, is multi-valued, with
                                  period
                                  > 2*pi*i, are you sure the correct value is always being taken by
                                  the software
                                  > you use?
                                  >

                                  Can you give some values? I would hope that Pari does this
                                  properly...

                                  I think it would be valuable if someone with better maths knowledge
                                  in this area than me could check through the paper

                                  38 #4421 10.41
                                  Fröberg, Carl-Erik
                                  On the prime zeta function.
                                  Nordisk Tidskr. Informationsbehandling (BIT) 8 1968 187--202.

                                  The function in the title is defined by the Dirichlet series
                                  $P(s)=\sum p\sp {-s}$, $s=\sigma+it$,
                                  $\sigma>1$, where the sum runs over all primes $p$.
                                  The author tabulates $P(s)$ for various values of
                                  $s$. Table I lists four roots of the equation $P(s)=0$


                                  which as far as I know is the only reference in print on this topic.
                                  Hopefully it will put my mind at ease as well!
                                  Using the mobius formula in Pari with Newton's method did find the
                                  zeros given in the paper, I won't be able to check until Tuesday
                                  what values they had for Re(z). All I did last year was check that
                                  the two prime zeta series did converge to the same value for complex
                                  values with R(z)>1. While I do seem to be finding zeros for the
                                  mobius series, it would be nice to know they are also zeros for
                                  the prime series. Perhaps David can elaborate some more on this?

                                  I think also one of the problems is that unlike the normal zeta
                                  function, the prime zeta function has only been minimally studied,
                                  so it would be nice to have moe info to check this is all above
                                  board!

                                  Andrew
                                • mikeoakes2@aol.com
                                  ... Yes, I can illustrate the problem. If the complex number z, in polar coordinates, is z = r*exp(i*theta), the usual definition of ln(z) is:- that function
                                  Message 16 of 18 , Jan 26, 2003
                                    Andrew Walker wrote:

                                    >
                                    > > I also have worries about the computation of ln(zeta(k*n)) when n
                                    > is not
                                    > > real: remembering that ln(z), for complex z, is multi-valued, with
                                    > period
                                    > > 2*pi*i, are you sure the correct value is always being taken by
                                    > the software
                                    > > you use?
                                    > >
                                    >
                                    > Can you give some values? I would hope that Pari does this
                                    > properly...
                                    >

                                    Yes, I can illustrate the problem.

                                    If the complex number z, in polar coordinates, is z = r*exp(i*theta), the
                                    usual definition of ln(z) is:-
                                    that function which is real for arguments on the real axis and continuous in
                                    the z-plane cut along the negative real axis, i.e. ln(z) = ln(r) + i*theta,
                                    with -2*pi <= theta <= 2*pi

                                    However, when z is complex, this definition does not respect one of the
                                    algebraic identities involved in going from the Euler product to the
                                    sum-of-series form of the zeta function, since it is not true in general that
                                    ln(z^s) = s*ln(z), even for real (and a fortiori for complex) exponents s.
                                    Example:-
                                    z = i, s = 5
                                    ln(z^s) = ln(i^5) = ln(i) = pi/2
                                    whereas
                                    s*ln(z) = 5*ln(z) = 5*(pi/2)

                                    The argument of ln(zeta(k*n)) certainly doesn't "respect" this cut along the
                                    negative real axis, taking as it does not only zero but even (as I pointed
                                    out before) even infinite values. So I claim that the ln() is a priori{Lat.}
                                    uncertain modulo{Lat.} (2*pi*i). If this uncertainty is not pinned down, your
                                    formal series may well still converge, but that fact alone does NOT prove
                                    that it converges to the "right" (in the sense of analytic continuation)
                                    values.

                                    For settling these formal questions of uniqueness and analytic continuation,
                                    appeals (pace{Lat.} David B.) to the convergence of Pari computations are
                                    irrelevant.

                                    Mike


                                    [Non-text portions of this message have been removed]
                                  • Jose Ramón Brox
                                    Hi, John: a) What is this program exactly about? b) Could you put a stop button on it? (It didn t want to be closed, then it tried to crash my operating system
                                    Message 17 of 18 , Feb 3, 2003
                                      Hi, John:

                                      a) What is this program exactly about?

                                      b) Could you put a stop button on it? (It didn't want to be closed, then it tried to crash my operating system :-P).

                                      Jose Brox
                                      ----- Original Message -----
                                      From: primenumbers@yahoogroups.com
                                      To: primenumbers@yahoogroups.com
                                      Sent: Monday, February 03, 2003 11:08 AM
                                      Subject: [PrimeNumbers] New file uploaded to primenumbers



                                      Hello,

                                      This email message is a notification to let you know that
                                      a file has been uploaded to the Files area of the primenumbers
                                      group.

                                      File : /Jon Perry's Stuff/randomprimewalkproject.zip
                                      Uploaded by : jon_perryuk <perry@...>
                                      Description : Random Prime Walks

                                      You can access this file at the URL

                                      http://groups.yahoo.com/group/primenumbers/files/Jon%20Perry%27s%20Stuff/randomprimewalkproject.zip

                                      To learn more about file sharing for your group, please visit

                                      http://help.yahoo.com/help/us/groups/files

                                      Regards,

                                      jon_perryuk <perry@...>






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                                      [Non-text portions of this message have been removed]
                                    • Jon Perry
                                      a)It s part of my prime squares research. I wondered (after staring at Ulam s Prime Spiral for about an hour), if any patterns might show themselves. For
                                      Message 18 of 18 , Feb 3, 2003
                                        a)It's part of my prime squares research. I wondered (after staring at
                                        Ulam's Prime Spiral for about an hour), if any patterns might show
                                        themselves. For example, with primes=2, at n=5000, the patterns look like
                                        smoke from a cigarette.

                                        b)I probably could, although I find Delphi threads quite difficult. Use
                                        Ctrl+Alt+Delete to safely close a program.

                                        Jon Perry
                                        perry@...
                                        http://www.users.globalnet.co.uk/~perry/maths/
                                        http://www.users.globalnet.co.uk/~perry/DIVMenu/
                                        BrainBench MVP for HTML and JavaScript
                                        http://www.brainbench.com
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