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Twin Conjecture visited

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  • Bryan Bartlett
    Disclaimer: I am a self taught amature mathematician. Information that is here that may have been already found is in no way me claiming responsibility for
    Message 1 of 12 , Mar 24, 2009
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      Disclaimer: I am a self taught amature mathematician. Information
      that is here that may have been already found is in no way me claiming
      responsibility for anothers work. Feel free to comment, or correct
      mistakes.. it is how i learn

      Let k be a number such that k-1 and k+1 are prime numbers (thus making
      k+1 and k-1 twin primes)

      the product of twin primes therefore will also always follow the format k^2-1

      Assume that k+1 is the nth prime, and k-1 is the n-1th prime.
      (k+1)# = p(1) * p(2) * p(3) * ... * p(n-2) * p(n-1) * p(n)

      since k-1 is the n-1th prime,

      (k-2)# = p(1) * p(2) * p(3) * ... * p(n-2)

      thus, (k+1)#/(k-2)# = p(n-1) * p(n)

      the products of twin primes can then also can be written as k+1# / k-2#

      therefore,

      (k+1)#/(k-2)# = k^2-1 for the twin primes (k-1), (k+1)

      this can be placed in a formula y=[(x+1)# / (x-2)#] -(x^2-1)

      If graphed, where the curve intercepts the x axis, it is the center
      mark of a twin prime

      From here, im looking to someone who can assist me in graphing this.
      I can manually enter the points, but i dont have a graphing program
      that can understand the Primorial expression #

      Also curious if this can be extended further to prove that as k
      approaches infinity, there are an infinite number of solutions where
      (k+1)#/(k-2)# = k^2-1

      --
      Bryan Bartlett
    • cino hilliard
      Hi Bryan, Interesting twin prime sieve. Using Pari in XP Pro, primorial(n) = local(p1, x); if(n==0||n==1,return(1));p1=1; forprime(x=2,n,p1*=x);return(p1)
      Message 2 of 12 , Mar 25, 2009
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        Hi Bryan,



        Interesting twin prime sieve.



        Using Pari in XP Pro,

        primorial(n) = local(p1, x); if(n==0||n==1,return(1));p1=1;

        forprime(x=2,n,p1*=x);return(p1)



        g1(x) = primorial(x+1)

        g2(x)=primorial(precprime(precprime(x)-1))
        g(n)=g1(n)/g2(n)-(n^2-1)

        ploth(k=1,200,g(k))



        To see some detail

        ploth(k=1,7,g(k))



        You can also do

        for(j=2,200,print(g(j)))



        Then Open Excel first and in the dos box, highlight copy and paste
        into excel to plot it there. If you don't open Excel first the clip board
        will be cleared by Excel start up. Just another MS pain unless of course
        there is a configuration option to disable this.



        Also Your Dos box properties must be in quick edit and insert mode for
        mouse highlight and right click to work. Also, layout screen buffer should be

        at least the number of lines you print.



        What is that curve?



        Enjoy,

        Cino Hilliard




        To: primenumbers@yahoogroups.com
        From: valareos@...
        Date: Wed, 25 Mar 2009 08:55:03 +1000
        Subject: [PrimeNumbers] Twin Conjecture visited





        Disclaimer: I am a self taught amature mathematician. Information
        that is here that may have been already found is in no way me claiming
        responsibility for anothers work. Feel free to comment, or correct
        mistakes.. it is how i learn

        Let k be a number such that k-1 and k+1 are prime numbers (thus making
        k+1 and k-1 twin primes)

        the product of twin primes therefore will also always follow the format k^2-1

        Assume that k+1 is the nth prime, and k-1 is the n-1th prime.
        (k+1)# = p(1) * p(2) * p(3) * ... * p(n-2) * p(n-1) * p(n)

        since k-1 is the n-1th prime,

        (k-2)# = p(1) * p(2) * p(3) * ... * p(n-2)

        thus, (k+1)#/(k-2)# = p(n-1) * p(n)

        the products of twin primes can then also can be written as k+1# / k-2#

        therefore,

        (k+1)#/(k-2)# = k^2-1 for the twin primes (k-1), (k+1)

        this can be placed in a formula y=[(x+1)# / (x-2)#] -(x^2-1)

        If graphed, where the curve intercepts the x axis, it is the center
        mark of a twin prime

        From here, im looking to someone who can assist me in graphing this.
        I can manually enter the points, but i dont have a graphing program
        that can understand the Primorial expression #

        Also curious if this can be extended further to prove that as k
        approaches infinity, there are an infinite number of solutions where
        (k+1)#/(k-2)# = k^2-1

        --
        Bryan Bartlett









        [Non-text portions of this message have been removed]
      • Maximilian Hasler
        ... To compute (k+1)# / (k-2)# it is completely inefficient to compute both primorials! (To get 999! / 998!, would you calculate both factorials ?) To compute
        Message 3 of 12 , Mar 25, 2009
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          On Wed, Mar 25, 2009 at 3:06 PM, cino hilliard wrote:
          >
          > primorial(n) = local(p1, x); if(n==0||n==1,return(1));p1=1;
          > forprime(x=2,n,p1*=x);return(p1)
          >
          > g1(x) = primorial(x+1)
          > g2(x)=primorial(precprime(precprime(x)-1))
          > g(n)=g1(n)/g2(n)-(n^2-1)

          To compute (k+1)# / (k-2)# it is completely inefficient to compute both primorials!
          (To get 999! / 998!, would you calculate both factorials ?)

          To compute (x+1)#/(x-2)# you could e.g. use

          pq(x)=local(pp=1);forprime(p=x-1,x+1, pp*=p);pp

          But this quotient equals 1 if there's no prime in [k-1,k+1],
          it equals k-1 or k or k+1 if this is a prime, but not the other two
          and it equals k²-1 if both k-1 and k+1 are prime.

          So the function g(n), suitably defined, does nothing else than to test primality of k-1 and/or k (if called with odd k) and/or k+1.
          As such, it is not a very innovative to find twin primes.

          It is always negative except in points corresponding to a twin prime, where it suddenly jumps to zero.

          Maximilian
        • David Broadhurst
          ... Indeed. But such postings about primorial/factorial tests of primality seem to be consecrated to *maximal* inefficiency. It seems that Wilson s primality
          Message 4 of 12 , Mar 25, 2009
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            --- In primenumbers@yahoogroups.com, "Maximilian Hasler"
            <maximilian.hasler@...> wrote:

            > To compute (k+1)# / (k-2)# it is completely inefficient
            > to compute both primorials!

            Indeed. But such postings about primorial/factorial tests of
            primality seem to be consecrated to *maximal* inefficiency.

            It seems that Wilson's primality test was not sufficiently
            inefficient for some people, who strive to do even worse?

            Given that strange motive, it seems appropriate to let folk
            amuse themselves, trying to outdo each other in greater
            inefficiency: chacun à son goût?

            David
          • Bryan Bartlett
            ... Good sir, This visitation on twin primes was not to FIND twin primes, but to create a formula base using a twin prime median ( the number between a twin
            Message 5 of 12 , Mar 26, 2009
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              > So the function g(n), suitably defined, does nothing else than to test
              > primality of k-1 and/or k (if called with odd k) and/or k+1.
              > As such, it is not a very innovative to find twin primes.
              >
              > It is always negative except in points corresponding to a twin prime, where
              > it suddenly jumps to zero.
              >

              Good sir,

              This visitation on twin primes was not to FIND twin primes, but to
              create a formula base using a twin prime median ( the number between a
              twin prime set) that holds true for all twin primes, and then
              therefore be also viewed in a graphical format to study its
              similarities to other, known linear formulas

              >To compute (k+1)# / (k-2)# it is completely inefficient to compute both primorials!
              >(To get 999! / 998!, would you calculate both factorials ?)

              I agree it is inefficient, however, it is a limitation of my own
              knowledge that compelled me not to simplify from there.

              simply put, as you say, 999!/998! i can easlly compute as 999, but
              that is because the order of numbers is known. Indeed, x!/(x-1)! will
              always equal x simply because order of numbers wont change depending
              on what x is.

              For primorials, that isnt necissarilly true. In the case of k + 1 and
              k - 1 being a twin prime, (k+1)# / (k-2)# will simplify into
              (k+1)(k-1). in fact, it will ONLY simplify to that if k + 1 and k -
              1 ARE both primes. It is this aspect that i have captured into the
              formula,

              another way to read it, is saying that (k+1)# / (k-2)# will always
              equal (k+1)(k-1) if, and only if, k+1 and k-1 are both primes.

              Thus, a formula can be graphed to show the relation.

              http://drecanis.ismywebsite.com/chart1.jpg

              as you can see, the larger a k is checked, the graph begins to take on
              the form of a parabolic function, with the occasional jumps to 0 due
              to being a twin prime median.

              Since the function does not reach a limitator, and indeed will
              streatch on to the infinite, one must also assume that the number of
              "jumps" is also infinite

              Thus, there must be an infinite number of twin prime medians, giving
              you also... an infinite number of twin primes... which ties this
              exersize back to.. the twin prime conjecture.




              Bryan
            • Maximilian Hasler
              ... Yes, but essentially the formula simplifies to g(x) ~ x^2 * (isprime(x-1) * isprime(x+1) - 1) The zeros of this function are at the same times its
              Message 6 of 12 , Mar 26, 2009
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                --- In primenumbers@yahoogroups.com, Bryan Bartlett wrote:
                >
                > > So the function g(n), suitably defined, does nothing else than to
                > > test primality of k-1 and/or k (if called with odd k) and/or k+1.
                > > It is always negative except in points corresponding to a twin
                > > prime, where it suddenly jumps to zero.
                >
                > Good sir,
                >
                > This visitation on twin primes was not to FIND twin primes, but to
                > create a formula base using a twin prime median (the number between
                > a twin prime set) that holds true for all twin primes, and then
                > therefore be also viewed in a graphical format to study its
                > similarities to other, known linear formulas


                Yes, but essentially the formula simplifies to

                g(x) ~ x^2 * (isprime(x-1) * isprime(x+1) - 1)

                The zeros of this function are at the same times its discontinuities, so even if we knew the complete graph of g except for the value in one single point, this would not give /any/ information about whether g is zero or not in this point;
                and in the same way, knowledge of g up to any value does not give any information about the existence or not of even a single zero beyond that point.


                > >To compute (k+1)# / (k-2)# it is completely inefficient to compute both primorials!
                > >(To get 999! / 998!, would you calculate both factorials ?)
                >
                > I agree it is inefficient, however, it is a limitation of my own
                > knowledge that compelled me not to simplify from there.

                You don't need to worry, since this was not a criticism of your idea but of Cino's PARI code, for which I gave a significant simplification.


                > Thus, a formula can be graphed to show the relation.
                >
                > http://drecanis.ismywebsite.com/chart1.jpg
                >
                > as you can see, the larger a k is checked, the graph begins to take on
                > the form of a parabolic function, with the occasional jumps to 0 due
                > to being a twin prime median.

                as I said

                > Since the function does not reach a limitator, and indeed will
                > streatch on to the infinite, one must also assume that the number of
                > "jumps" is also infinite

                No, one must not since one can't.
                This reasoning has not any logical foundation: Consider the function x² which also "does not reach a limitator and will stretch to the infinite", but it does not have ANY jumps.

                Maximilian
              • cino hilliard
                ... I read the origional post as a request for a primorial function and a graph. My intent was to stick as closely as possible to the poster s origional
                Message 7 of 12 , Mar 26, 2009
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                  Maximilian wrote:
                  >You don't need to worry, since this was not a criticism of your idea but of

                  >Cino's PARI code, for which I gave a significant simplification.


                  Org post:
                  >>this can be placed in a formula y=[(x+1)# / (x-2)#] -(x^2-1)
                  >>If graphed, where the curve intercepts the x axis, it is the center
                  >>mark of a twin prime

                  >>From here, im looking to someone who can assist me in graphing this.
                  >>I can manually enter the points, but i dont have a graphing program
                  >>that can understand the Primorial expression #

                  I read the origional post as a request for a primorial function and
                  a graph. My intent was to stick as closely as possible to the poster's
                  origional formula. I did not read it as an optimization request.

                  So I guess I failed the efficiency test. So what?

                  Here are some excerpts from the Texas Instruments SR-56 programable
                  calculator manual that ring true today. A difference is we have a lot
                  more than 100 steps to do programs today.



                  DEVELOPMENT OF PROGRAMMING STYLE

                  There is no single correct programming solution to a problem. Just as
                  no two writers use exactly the same words to describe the same thing,
                  no two programmers use exactly the same instruction sequences to solve
                  a given problem.

                  ...

                  As you gain experience in programming the SR-56, you will develope your own
                  unique style. That style may become one of incredible craftiness and ingenuity
                  which makes frequent use of all the instructions available. Or it might become
                  one of conservative and straight forward coding, using primarily the more basic
                  instructions, taking up more memory space, but so clear in purpose that program
                  operation can easily be discerned simply by inspecting the code.



                  OPTIMIZING A PROGRAM



                  Condensing a program to a smaller number of steps is a time consuming exercise.
                  If a program is less than 100 steps and operates properly, any time spent to condense
                  the program, in most cases, is unnecessary except for the personal satisfaction of doing
                  it.


                  This is not a bad way to think while writing code no matter what language you use.

                  Moreover, it is also applicable to prime numbers, formulas, and numerous other
                  cognitive diciplines.



                  Enjoy,

                  Cino



                  [Non-text portions of this message have been removed]
                • Maximilian Hasler
                  Dear Cino, sorry if my post sounded a bit too negative... Above all, I wanted to point out that the basic idea, ... is somehow misleading ; IMHO a curve which
                  Message 8 of 12 , Mar 26, 2009
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                    Dear Cino,

                    sorry if my post sounded a bit too negative...

                    Above all, I wanted to point out that the basic idea,

                    >>>If graphed, where the curve intercepts the x axis, it is the center

                    is somehow misleading ; IMHO a curve which is far from the axis except for some particular isolated points where "suddenly / jump-wise" f(x)=0, cannot really be said to "intercept" the axis in the usual understanding, and (as mentioned earlier) even less be used to predict further zeros.

                    Regarding the optimisation issue, I completely agree with what you and the TI manual say ; while I plead guilty of having the bad habit to always optimize too much (I just can't help!), I don't consider this as a virtue at all.

                    OTOH, even if one *writes* expressions like n! / k!,
                    one does not *mean* this, but one does mean (k+1)*...*n.
                    At least I did understand the

                    > Org post:
                    > >>this can be placed in a formula y=[(x+1)# / (x-2)#] -(x^2-1)

                    in that way.
                    Therefore I don't think that writing the fraction as
                    product of the primes p >= x-1, p <= x+1,
                    is an obscure optimization - and it takes less time to write this than to define the primorial function and the quotient of the two function calls thereof.

                    Considering the OP's problem, the isprime() function is more elementary and easier to implement than primorial() using the forprime() loop (e.g. in excel or whatever graphing proggy that does not have it). That's why I proposed my simplification.

                    Sorry again if I did formulate it in an inappropriately critical way...
                    I hope we now can "make peace" and turn to other challenges...
                    Regards,
                    Maximilian
                  • Bryan Bartlett
                    Maximilian and Cino, You both have been extremly helpful, and I appreciate it greatly. as a point for both of you. As a work in progress for any math in
                    Message 9 of 12 , Mar 26, 2009
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                      Maximilian and Cino,

                      You both have been extremly helpful, and I appreciate it greatly.

                      as a point for both of you.

                      As a work in progress for any math in progress, I do like to have any
                      program i use keepot as close as possible to my formulas. it makes it
                      easier to look for mistakes. A simplified method would work fine for
                      the program, but a math formula doesnt follow the same logic codes.

                      On a counter note, A finished math formula that can be simplified in a
                      programming language is a helpful boon. It allows one to condense
                      several manual computations into a simpler form.



                      Now as far as the formula, as it stands, not able to prove the twin
                      conjecture as of yet, I do agree. I do not agree, however, that it is
                      impossible to do, with this, or any other formula.

                      A few interesting tidbits... if you set the answer to my equation to
                      0 (thus making the equation a twin prime) and then solved for x, the
                      entire equation collapses in on itself, to get an identity answer ( 1
                      = 1, 0=0, x=x, for examle)

                      now, correct me if im wrong, but i was taught long ago in high school,
                      that if any equation collapses onto itself, then that equation has an
                      infinte number of solutions.

                      Case in point, given the equation y = xy

                      logical math will say the answer is simple, x = 1 for any value of y
                      (divide both sides by y).

                      However, if asked to solve for x if y is 0, the equation collapes down
                      to 0=0.. in other words, if Y is 0, there is an infinite number of x's
                      that can be the answer.

                      in my equation, if the answer is 0, then the variable must be a twin
                      prime median. Yet, when i set the equation to 0 (thus to find a
                      formula for finding twin primes, as was my original goal), the entire
                      system collapses into 0=0

                      Im not quite happy with this result, because it also could mean that i
                      used my formula to justify my formula, which is a big no no.


                      So, a challenge for you two.

                      If you accept my thought that if (k+1)#/(k-2)#-(k^2-1) is 0, then k is
                      a twin prime median by my OP thought process, then figure out the
                      following

                      assume k is a twin prime median. Find all number's p such that k+p
                      is also a twin prime median

                      whether you get an answer that there are infinite values for p, finite
                      values of p, or even that it can not be calculated from my formula, id
                      like to see the proof.

                      Also, I got my own formula based on K+P, and got a very sneaky
                      suspicion the "jumps" will disappear and fall into a smoother curve
                      structure. However, the formula will create a 3d structre. -.- can
                      pari do 3d graphs?

                      in other words.. I am curious to see if the irregular function i
                      created is merely a flat representation of a 3d structure.
                    • Bryan Bartlett
                      Adding to this study ive been conducting, i decided to see if I could find a way to simplify my formula, but keep the original conjecture alive, to perhaps get
                      Message 10 of 12 , Mar 28, 2009
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                        Adding to this study ive been conducting, i decided to see if I could
                        find a way to simplify my formula, but keep the original conjecture
                        alive, to perhaps get rid of the "spikes" my formulas will graph

                        I then made this assumtion... let m be a number such that m+n is a
                        twin prime median

                        This allows me to revisit my original formula, and repace k with m+n
                        to recieve the following

                        (m + n)^2 - 1

                        This is the result of the product of the twin prime m+n+1,m+n-1

                        I then simplified the second side to simply P(n) * P(n-1).. which is
                        simply the product of a prime, and the next lowest prime.

                        Now, setting (m + n)^2 - 1 = P(n)* P(n-1) hase an interesting effect.
                        for every value of n, there can be two values of m to make this
                        formula true (which is based off of my original twin prime median
                        conjecture)

                        now, obviously this formula, as is, does not have the same effect as
                        the one which uses
                        (k+1)#/(k-2)#, because for non twin prime medians m+n, m is an
                        irrational number in the form of -n±√[{P(n) * P(n-1)}+1]

                        However, for where m+n is a twin prime median, m is a whole number

                        Thus, a graph of m = -n±√[{P(n) * P(n-1)}+1] will show m+n as a twin
                        prime median where m and n is a whole number.

                        will play with pari to see if i can get this graphed.

                        But this formula, as written, or at least manipulated a bit more,
                        should be able to prove, or disprove the twin prime conjecture.

                        Is there a way of proving with the new formula, that as n approaches
                        infinity, the amount of whole number answers for m also approaches
                        infinity?
                      • Bryan Bartlett
                        Been working on this, and i have made a great stride to helping prove the twin conjecture. Here is what i have so far. Let n be an integer equal or greater
                        Message 11 of 12 , May 12, 2009
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                          Been working on this, and i have made a great stride to helping prove
                          the twin conjecture.

                          Here is what i have so far.

                          Let n be an integer equal or greater than 2.

                          for every number n, there is a number m such that

                          m=-n+√[{P(n) * P(n-1)}+1]

                          if m is also an integer, then m+n is a twin prime median, such that
                          m+n+1 and m+n-1 is a twin prime

                          for those using Pari, the data can be retrieved using the following commands

                          f(n)= -n+sqrt((prime(n)*prime(n-1))+1)
                          for(k=2,200,print(f(k)))

                          Last bit to prove... that there is an infinite number of solutions
                          where m and n are integers.

                          Anyone want to help, feel free to jump in

                          2009/3/28 Bryan Bartlett <valareos@...>:
                          > Adding to this study ive been conducting, i decided to see if I could
                          > find a way to simplify my formula, but keep the original conjecture
                          > alive, to perhaps get rid of the "spikes" my formulas will graph
                          >
                          > I then made this assumtion... let m be a number such that m+n is a
                          > twin prime median
                          >
                          > This allows me to revisit my original formula, and repace k with m+n
                          > to recieve the following
                          >
                          > (m + n)^2 - 1
                          >
                          > This is the result of the product of the twin prime m+n+1,m+n-1
                          >
                          > I then simplified the second side to simply P(n) * P(n-1).. which is
                          > simply the product of a prime, and the next lowest prime.
                          >
                          > Now, setting (m + n)^2 - 1 = P(n)* P(n-1) hase an interesting effect.
                          > for every value of n, there can be two values of m to make this
                          > formula true (which is based off of my original twin prime median
                          > conjecture)
                          >
                          > now, obviously this formula, as is, does not have the same effect as
                          > the one which uses
                          > (k+1)#/(k-2)#, because for non twin prime medians m+n, m is an
                          > irrational number in the form of -n±√[{P(n) * P(n-1)}+1]
                          >
                          > However, for where m+n is a twin prime median, m is a whole number
                          >
                          > Thus, a graph of m = -n±√[{P(n) * P(n-1)}+1] will show m+n as a twin
                          > prime median where m and n is a whole number.
                          >
                          > will play with pari to see if i can get this graphed.
                          >
                          > But this formula, as written, or at least manipulated a bit more,
                          > should be able to prove, or disprove the twin prime conjecture.
                          >
                          > Is there a way of proving with the new formula, that as n approaches
                          > infinity, the amount of whole number answers for m also approaches
                          > infinity?
                          >



                          --
                          Bryan Bartlett
                        • Bryan Bartlett
                          ... But unlike my previous attempts, this one doesnt need P(n), P(n-1) to be a twin prime. Thus, that part falls under the proof of infinite primes now, and
                          Message 12 of 12 , May 12, 2009
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                            2009/5/13 Maximilian Hasler <maximilian.hasler@...>:

                            > the only problem is the P(n)*P(n-1) part.


                            But unlike my previous attempts, this one doesnt need P(n), P(n-1) to
                            be a twin prime.

                            Thus, that part falls under the proof of infinite primes now, and not
                            under a twin prime conjecture

                            This gets rid of my problem of using theory to solve a theory.

                            Also, the graph no longer has the spikes in it, in fact, it resembles
                            a sine wave/parabolic curve addition

                            so to restate what i said above...

                            Prove of an infinite number of solutions to the formula m=-n+√[{P(n) *
                            P(n-1)}+1] is done.

                            if a proof of inifinte number of solutions where m is an integer (n
                            has to be an integer since it is used also in the order of primes) can
                            be made, then there can be also said to be an infinite number of twin
                            primes.

                            --
                            Bryan Bartlett
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