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Factorization of 6 consecutive 1803-digit integers

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  • David Broadhurst
    Let b=4011209802600 m=16812956160*b x=5000001251617*b y=(2*x^6+3*x^3-148)^2 N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4 then N+k is
    Message 1 of 16 , Sep 7, 2007
      Let

      b=4011209802600
      m=16812956160*b
      x=5000001251617*b
      y=(2*x^6+3*x^3-148)^2
      N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4

      then N+k is factorized for k in [0,5].

      Proof:
      http://physics.open.ac.uk/~dbroadhu/cert/ifac6nfs.zip

      Software used:
      GGNFS (with SNFS difficulty = 152.1)
      GMP-ECM
      Msieve
      OpenPFGW
      Pari-GP
      Primo

      Neighbouring composites:
      c1758=(N-1)/(5*239283371994839*543315317127619*4154925272246149)
      c1771=(N+6)/(2*8081*33697745047039*268690037677741)
      subjected to 400 ECM curves at B1=250000

      David Broadhurst
      7 September 2007
    • Jens Kruse Andersen
      ... Congratulations on your 10th record in a month! (counting as multiple records when it was record for multiple lengths) I m impressed. The record page is
      Message 2 of 16 , Sep 7, 2007
        David Broadhurst wrote:
        > Let
        >
        > b=4011209802600
        > m=16812956160*b
        > x=5000001251617*b
        > y=(2*x^6+3*x^3-148)^2
        > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
        >
        > then N+k is factorized for k in [0,5].

        Congratulations on your 10th record in a month!
        (counting as multiple records when it was record for multiple lengths)
        I'm impressed.

        The record page is updated (people can probably find the link by now after I
        have spammed the list with it for a month).

        --
        Jens Kruse Andersen
      • David Broadhurst
        ... It is unlikely that this 72nd order Ansatz will be easily improved upon, since it seems fully to exploit the best known solution to the
        Message 3 of 16 , Sep 8, 2007
          --- In primeform@yahoogroups.com, "David Broadhurst"
          <d.broadhurst@...> wrote:

          > b=4011209802600
          > m=16812956160*b
          > x=5000001251617*b
          > y=(2*x^6+3*x^3-148)^2
          > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
          >
          > then N+k is factorized for k in [0,5].

          It is unlikely that this 72nd order Ansatz will be easily
          improved upon, since it seems fully to exploit the best
          known solution to the Prouhet-Tarry-Escott problem:

          http://euler.free.fr/eslp/TarryPrb.htm
          http://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html

          David
        • David Broadhurst
          ... PS: I remark that the search reported in http://portal.acm.org/citation.cfm?id=950414.950443 failed to improve on Nuutti Kuosa s wonderful discovery of 3
          Message 4 of 16 , Sep 8, 2007
            --- In primeform@yahoogroups.com, "David Broadhurst"
            <d.broadhurst@...> wrote:

            > It is unlikely that this 72nd order Ansatz will be easily
            > improved upon, since it seems fully to exploit the best
            > known solution to the Prouhet-Tarry-Escott problem

            PS: I remark that the search reported in

            http://portal.acm.org/citation.cfm?id=950414.950443

            failed to improve on Nuutti Kuosa's wonderful discovery of 3 Sep 1999:

            http://euler.free.fr/eslp/k246810.htm

            David
          • David Broadhurst
            ... I m even more impressed by that fact that Peter Borwein, Petr Lisonek and Colin Percival were unable to improve on Nuutti s polynomial, after expending
            Message 5 of 16 , Sep 8, 2007
              --- In primeform@yahoogroups.com, "Jens Kruse Andersen"
              <jens.k.a@...> wrote:

              > Congratulations on your 10th record in a month!
              > I'm impressed.

              I'm even more impressed by that fact that
              Peter Borwein, Petr Lisonek and Colin Percival
              were unable to improve on Nuutti's polynomial,
              after expending 10^17 floating point operations.

              Such Diophantine problems are amongst the weirdest that we know.

              David
            • David Broadhurst
              ... Their Math.Comp. article currently leaks out of googling 35, 47, 94, 121, 146, 148 even though it /ought/ to be password protected. Verbum sat sapienti?
              Message 6 of 16 , Sep 8, 2007
                --- In primeform@yahoogroups.com, "David Broadhurst"
                <d.broadhurst@...> wrote:

                > Peter Borwein, Petr Lisonek and Colin Percival
                > were unable to improve on Nuutti's polynomial,
                > after expending 10^17 floating point operations.

                Their Math.Comp. article currently leaks out of googling

                "35, 47, 94, 121, 146, 148"

                even though it /ought/ to be password protected.

                Verbum sat sapienti?

                David
              • Pierre CAMI
                I just need to understand , I found that n^14-n^7+1 is never prime for n integer until more that 6000000 If this is always true , why ? When using pfgw
                Message 7 of 16 , Sep 13, 2007
                  I just need to understand , I found that n^14-n^7+1 is never prime for n
                  integer until more that 6000000
                  If this is always true , why ?
                  When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
                  but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
                  factor found ! What is the bug for PRP answer ?
                  Thanks for help
                  Pierre
                • Sean A. Irvine
                  Yes, it is always true because you can factor the polynomial n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1) S.
                  Message 8 of 16 , Sep 13, 2007
                    Yes, it is always true because you can factor the polynomial

                    n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

                    S.

                    Pierre CAMI wrote:
                    > I just need to understand , I found that n^14-n^7+1 is never prime for n
                    > integer until more that 6000000
                    > If this is always true , why ?
                    > When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
                    > but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
                    > factor found ! What is the bug for PRP answer ?
                    > Thanks for help
                    > Pierre
                    >
                    >
                    >
                    >
                    > List address:
                    > Post message: primeform@yahoogroups.com
                    > Web access:
                    > http://groups.yahoo.com/group/primeform
                    > Yahoo! Groups Links
                    >
                    >
                    >
                    >
                  • Jens Kruse Andersen
                    ... n^14-n^7+1 = (n^2-n+1)*(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1) ... 729^14-729^7+1 is indeed 3-PRP. Note that 729 = 3^6. Some number forms involving powers of b
                    Message 9 of 16 , Sep 13, 2007
                      Pierre CAMI wrote:
                      > I found that n^14-n^7+1 is never prime

                      n^14-n^7+1 = (n^2-n+1)*(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

                      > When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
                      > but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
                      > factor found ! What is the bug for PRP answer ?

                      729^14-729^7+1 is indeed 3-PRP. Note that 729 = 3^6.
                      Some number forms involving powers of b are b-PRP.
                      Base 3 is default in pfgw. Here is base 2:

                      C:\Users\Jens>pfgw -b2 -q729^^14-729^^7+1
                      PFGW Version 1.2.0 for Windows [FFT v23.8]

                      Switching to Exponentiating using GMP
                      729^14-729^7+1 is composite: [D66D28B9C614392] (0.0004s+0.0003s)

                      --
                      Jens Kruse Andersen
                    • David Broadhurst
                      ... wrote (as Jens had done before) ... The larger factor is Phi(42,n), which is a BPSW probable prime for 28,536 of the first million natural numbers: ct=0;
                      Message 10 of 16 , Sep 13, 2007
                        --- In primeform@yahoogroups.com, "Sean A. Irvine" <sairvin@...>
                        wrote (as Jens had done before)

                        > n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

                        The larger factor is Phi(42,n), which is a BPSW probable prime
                        for 28,536 of the first million natural numbers:

                        ct=0;
                        for(n=1,10^6,if(ispseudoprime(subst(polcyclo(42),x,n)),ct=ct+1));
                        print(ct)

                        28536

                        David
                      • David Broadhurst
                        ... Today I made some progress with the PTE problem: http://listserv.nodak.edu/cgi-bin/wa.exeA2=ind0709&L=nmbrthry&P=R994&D ... It may take a good while longer
                        Message 11 of 16 , Sep 26, 2007
                          --- In primeform@yahoogroups.com, "David Broadhurst"
                          <d.broadhurst@...> wrote:

                          > It is unlikely that this 72nd order Ansatz will be easily
                          > improved upon, since it seems fully to exploit the best
                          > known solution to the Prouhet-Tarry-Escott problem:
                          > http://euler.free.fr/eslp/TarryPrb.htm
                          > http://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html

                          Today I made some progress with the PTE problem:

                          http://listserv.nodak.edu/cgi-bin/wa.exeA2=ind0709&L=nmbrthry&P=R994&D
                          > Perhaps a CRT strategy might lead to a solution with degree s > 12.

                          It may take a good while longer to see if this feeds into Jens's
                          consecutive-factorization page, but at least it is pleasing to find
                          a new PTE solution, rather economically, using 1700-year-old
                          mathematics.

                          David
                        • David Broadhurst
                          ... but either Yahoo or I mangled the link, which should have been http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0709&L=nmbrthry&P=R994 In any case, the
                          Message 12 of 16 , Sep 26, 2007
                            --- In primeform@yahoogroups.com, "David Broadhurst"
                            <d.broadhurst@...> wrote:

                            > Today I made some progress with the PTE problem:

                            but either Yahoo or I mangled the link, which should have been

                            http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0709&L=nmbrthry&P=R994

                            In any case, the sentiment

                            > It may take a good while longer to see if this feeds into Jens's
                            > consecutive-factorization page, but at least it is pleasing to find
                            > a new PTE solution, rather economically, using 1700-year-old
                            > mathematics.

                            still stands.

                            David
                          • Andrey Kulsha
                            ... The link is http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm but it doesn t work anymore... Andrey
                            Message 13 of 16 , May 27, 2010
                              Jens Kruse Andersen wrote:
                              > David Broadhurst wrote:
                              > > Let
                              > >
                              > > b=4011209802600
                              > > m=16812956160*b
                              > > x=5000001251617*b
                              > > y=(2*x^6+3*x^3-148)^2
                              > > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
                              > >
                              > > then N+k is factorized for k in [0,5].
                              >
                              > Congratulations on your 10th record in a month!
                              > (counting as multiple records when it was record for multiple lengths)
                              > I'm impressed.
                              >
                              > The record page is updated (people can probably find the link by now after
                              > I have spammed the list with it for a month).

                              The link is

                              http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm

                              but it doesn't work anymore...

                              Andrey
                            • Norman Luhn
                              And this ? http://users.cybercity.dk/~dsl522332/ ... Von: Andrey Kulsha Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit
                              Message 14 of 16 , May 27, 2010
                                And this ?

                                http://users.cybercity.dk/~dsl522332/

                                --- Andrey Kulsha <Andrey_601@...> schrieb am Do, 27.5.2010:

                                Von: Andrey Kulsha <Andrey_601@...>
                                Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit integers
                                An: primeform@yahoogroups.com
                                Datum: Donnerstag, 27. Mai, 2010 14:35 Uhr







                                 









                                Jens Kruse Andersen wrote:

                                > David Broadhurst wrote:

                                > > Let

                                > >

                                > > b=4011209802600

                                > > m=16812956160*b

                                > > x=5000001251617*b

                                > > y=(2*x^6+3*x^3-148)^2

                                > > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4

                                > >

                                > > then N+k is factorized for k in [0,5].

                                >

                                > Congratulations on your 10th record in a month!

                                > (counting as multiple records when it was record for multiple lengths)

                                > I'm impressed.

                                >

                                > The record page is updated (people can probably find the link by now after

                                > I have spammed the list with it for a month).



                                The link is



                                http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm



                                but it doesn't work anymore...



                                Andrey
























                                [Non-text portions of this message have been removed]
                              • Andrey Kulsha
                                ... Thanks Norman. There s also a record for k=20: http://tech.groups.yahoo.com/group/primeform/message/9982 But it doesn t fit the tables unfortunately. Best
                                Message 15 of 16 , May 27, 2010
                                  > And this ?
                                  >
                                  > http://users.cybercity.dk/~dsl522332/

                                  Thanks Norman.

                                  There's also a record for k=20:
                                  http://tech.groups.yahoo.com/group/primeform/message/9982

                                  But it doesn't fit the tables unfortunately.

                                  Best regards,

                                  Andrey
                                • Norman Luhn
                                  Maybe here ? http://www.math.uni.wroc.pl/~jwr/cons-fac/ ... Von: Andrey Kulsha Betreff: Re: [primeform] Factorization of 6 consecutive
                                  Message 16 of 16 , May 27, 2010
                                    Maybe here ?
                                    http://www.math.uni.wroc.pl/~jwr/cons-fac/


                                    --- Andrey Kulsha <Andrey_601@...> schrieb am Do, 27.5.2010:

                                    Von: Andrey Kulsha <Andrey_601@...>
                                    Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit integers
                                    An: primeform@yahoogroups.com
                                    Datum: Donnerstag, 27. Mai, 2010 15:18 Uhr







                                     









                                    > And this ?

                                    >

                                    > http://users.cybercity.dk/~dsl522332/



                                    Thanks Norman.



                                    There's also a record for k=20:

                                    http://tech.groups.yahoo.com/group/primeform/message/9982



                                    But it doesn't fit the tables unfortunately.



                                    Best regards,



                                    Andrey
























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