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13 consecutive factorizations at 500 digits

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  • Joe
    I d like to report 13 consecutive factorizations at 500 digits. ... N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2
    Message 1 of 16 , Dec 15, 2009
      I'd like to report 13 consecutive factorizations at 500 digits.
      ------------------------------
      N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
      where x = 2^92+227683166;

      N-k is factored for k={0..12}
      ------------------------------

      All primes were proved using PARI/GP.

      PARI/GP verification script available here:
      http://immortaltheory.com/cnt/verify_k13.gp

      As always my brother John Michael Crump helped with his ECM support! :)

      Factorizations are abbreviated as follows:
      ---------------------------------------------
      N-0 = 2^6 * 3^2 * 41^2 * 59^4 * 199^2 * 3907^2 * 1067282959^2 * p59^2 * p67^2 * p104^2
      ---------------------------------------------
      N-1 = 5 * 7^2 * 19 * 2039 * 2203 * 18917 * 20773 * 128663 * 206197 * 337283 * 4205437 * 4253763823 * 10361106142511 * 48219445919633 * 38944767299226517 * 823146395611037243 * 2567947475640367903 * p31 * p46 * p105 * p189
      ---------------------------------------------
      N-2 = 2 * 31 * 71 * 479 * 601 * 2521 * 10079 * 11351 * p34 * p35 * p43 * p46 * p94 * p112 * p119
      ---------------------------------------------
      N-3 = 3 * 61 * 1019 * 14243 * 52883 * 269749 * 73255439 * 1462827239 * p22 * p37 * p44 * p90 * p125 * p147
      ---------------------------------------------
      N-4 = 2^2 * 829 * 26342725653094583 * p28 * p33 * p53 * p63 * p88 * p217
      ---------------------------------------------
      N-5 = 4001 * 33739 * p491
      ---------------------------------------------
      N-6 = 2 * 3 * 5^2 * 29 * 2664932713 * 64053375527497 * p35 * p122 * p317
      ---------------------------------------------
      N-7 = p500
      ---------------------------------------------
      N-8 = 2^3 * 7 * 103 * 167 * 1823 * 9436961405541853927 * p67 * p75 * p330
      ---------------------------------------------
      N-9 = 3^5 * 11 * 37 * 73 * 89 * 1697 * 22111 * 24379 * 50993 * 75403 * 294431 * 791899 * 7143163 * 37162403 * 40238771 * 318066373 * 774022187 * 1420502627 * 319986975727 * 360802459587989 * 4907888847800929 * p28 * p40 * p43 * p93 * p166
      ---------------------------------------------
      N-10 = 2 * 13 * 227 * 1889 * p493
      ---------------------------------------------
      N-11 = 5 * 5261 * p23 * p473
      ---------------------------------------------
      N-12 = 2^2 * 3 * 1022689 * 11589211679 * p482
      ------------------------------

      The ends, N+1 and N-13, were subjected to ~1700 ECM curves at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.

      PS: With another candidate x, I came within a c233 of finding a 14-streak. Unfortunately though that
      one didn't factor after a considerable ECM effort. But suggests longer streaks are within range
      with time and/or luck.
    • Jaroslaw Wroblewski
      CONGRATULATIONS !!! Impressive! What more can I say? My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated (since the results still fits Jens s limit
      Message 2 of 16 , Dec 15, 2009
        CONGRATULATIONS !!!

        Impressive!

        What more can I say?

        My site
        http://www.math.uni.wroc.pl/~jwr/cons-fac/
        is updated (since the results still fits Jens's limit of 500 digits,
        it will be listed primarily at his site).

        Jarek

        2009/12/16 Joe <joecr@...>
        >
        >
        >
        > I'd like to report 13 consecutive factorizations at 500 digits.
        > ------------------------------
        > N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
        > where x = 2^92+227683166;
        >
        > N-k is factored for k={0..12}
        > ------------------------------
        >
        > All primes were proved using PARI/GP.
        >
        > PARI/GP verification script available here:
        > http://immortaltheory.com/cnt/verify_k13.gp
        >
        > As always my brother John Michael Crump helped with his ECM support! :)
        >
        > Factorizations are abbreviated as follows:
        > ---------------------------------------------
        > N-0 = 2^6 * 3^2 * 41^2 * 59^4 * 199^2 * 3907^2 * 1067282959^2 * p59^2 * p67^2 * p104^2
        > ---------------------------------------------
        > N-1 = 5 * 7^2 * 19 * 2039 * 2203 * 18917 * 20773 * 128663 * 206197 * 337283 * 4205437 * 4253763823 * 10361106142511 * 48219445919633 * 38944767299226517 * 823146395611037243 * 2567947475640367903 * p31 * p46 * p105 * p189
        > ---------------------------------------------
        > N-2 = 2 * 31 * 71 * 479 * 601 * 2521 * 10079 * 11351 * p34 * p35 * p43 * p46 * p94 * p112 * p119
        > ---------------------------------------------
        > N-3 = 3 * 61 * 1019 * 14243 * 52883 * 269749 * 73255439 * 1462827239 * p22 * p37 * p44 * p90 * p125 * p147
        > ---------------------------------------------
        > N-4 = 2^2 * 829 * 26342725653094583 * p28 * p33 * p53 * p63 * p88 * p217
        > ---------------------------------------------
        > N-5 = 4001 * 33739 * p491
        > ---------------------------------------------
        > N-6 = 2 * 3 * 5^2 * 29 * 2664932713 * 64053375527497 * p35 * p122 * p317
        > ---------------------------------------------
        > N-7 = p500
        > ---------------------------------------------
        > N-8 = 2^3 * 7 * 103 * 167 * 1823 * 9436961405541853927 * p67 * p75 * p330
        > ---------------------------------------------
        > N-9 = 3^5 * 11 * 37 * 73 * 89 * 1697 * 22111 * 24379 * 50993 * 75403 * 294431 * 791899 * 7143163 * 37162403 * 40238771 * 318066373 * 774022187 * 1420502627 * 319986975727 * 360802459587989 * 4907888847800929 * p28 * p40 * p43 * p93 * p166
        > ---------------------------------------------
        > N-10 = 2 * 13 * 227 * 1889 * p493
        > ---------------------------------------------
        > N-11 = 5 * 5261 * p23 * p473
        > ---------------------------------------------
        > N-12 = 2^2 * 3 * 1022689 * 11589211679 * p482
        > ------------------------------
        >
        > The ends, N+1 and N-13, were subjected to ~1700 ECM curves at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.
        >
        > PS: With another candidate x, I came within a c233 of finding a 14-streak. Unfortunately though that
        > one didn't factor after a considerable ECM effort. But suggests longer streaks are within range
        > with time and/or luck.
        >
        >
      • Jens Kruse Andersen
        ... Congratulations on the first 13! I m impressed. The length record has now increased by 3 in 18 days.
        Message 3 of 16 , Dec 16, 2009
          Joe wrote:
          > I'd like to report 13 consecutive factorizations at 500 digits.
          > ------------------------------
          > N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4
          > + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
          > where x = 2^92+227683166;
          >
          > N-k is factored for k={0..12}

          Congratulations on the first 13!
          I'm impressed.
          The length record has now increased by 3 in 18 days.
          http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
          is updated.

          > The ends, N+1 and N-13, were subjected to ~1700 ECM curves
          > at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.

          You didn't mention the prime factors but it was easy to find:
          N+1 = 17 * c498
          N-13 = 7909460491591 * 854342030873171 * c472

          --
          Jens Kruse Andersen
        • djbroadhurst
          ... thus neatly satisfying both Jens and Jarek. ... 17, perhaps? {N=(2*x^3-3*x^2-5*x+2)^2 *(4*x^6-12*x^5-11*x^4+38*x^3+13*x^2-20*x-6)^2/16;}
          Message 4 of 16 , Dec 16, 2009
            --- In primeform@yahoogroups.com,
            "Joe" <joecr@...> wrote:

            > 13 consecutive factorizations at 500 digits

            thus neatly satisfying both Jens and Jarek.

            > longer streaks are within range

            17, perhaps?

            {N=(2*x^3-3*x^2-5*x+2)^2
            *(4*x^6-12*x^5-11*x^4+38*x^3+13*x^2-20*x-6)^2/16;}
            {for(k=0,16,f=factor(N-k)[,1];if(#f>1,
            print([k,vector(#f,j,poldegree(f[j]))])))}
            [0, [3, 6]]
            [1, [3, 6, 9]]
            [2, [2, 4, 6, 6]]
            [3, [6, 6, 6]]
            [4, [3, 6, 9]]
            [6, [6, 12]]
            [8, [6, 12]]
            [9, [1, 1, 1, 3, 6, 6]]
            [14, [6, 12]]
            [16, [9, 9]]

            All the best!

            David
          • djbroadhurst
            Factorization of 16 consecutive 303-digit integers Let x = 9*(2^123 + 23815893) N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 then for k = 0
            Message 5 of 16 , Dec 23, 2009
              Factorization of 16 consecutive 303-digit integers

              Let x = 9*(2^123 + 23815893)
              N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216
              then for k = 0 to 15 the factorization of N+k is in
              http://physics.open.ac.uk/~dbroadhu/cert/ifac16.zip
              I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
              It is unlikely that either of the composite integers
              (N-1)/(3*5*107614765903*707599683401162722538492715491)
              (N+16)/(2*59*347*509*28099*71147*63442396558253*535991020920557)
              is divisible by a prime less than 10^35.

              David Broadhurst, 24 December, 2009
            • Jaroslaw Wroblewski
              Congratulations!!! My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Jarek 2009/12/24 djbroadhurst ... [Non-text
              Message 6 of 16 , Dec 23, 2009
                Congratulations!!!

                My site
                http://www.math.uni.wroc.pl/~jwr/cons-fac/
                is updated.

                Jarek

                2009/12/24 djbroadhurst <d.broadhurst@...>

                >
                >
                >
                >
                > Factorization of 16 consecutive 303-digit integers
                >
                > Let x = 9*(2^123 + 23815893)
                > N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216
                > then for k = 0 to 15 the factorization of N+k is in
                > http://physics.open.ac.uk/~dbroadhu/cert/ifac16.zip<http://physics.open.ac.uk/%7Edbroadhu/cert/ifac16.zip>
                > I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
                > It is unlikely that either of the composite integers
                > (N-1)/(3*5*107614765903*707599683401162722538492715491)
                > (N+16)/(2*59*347*509*28099*71147*63442396558253*535991020920557)
                > is divisible by a prime less than 10^35.
                >
                > David Broadhurst, 24 December, 2009
                >
                >
                >


                [Non-text portions of this message have been removed]
              • djbroadhurst
                ... You might add that N-1 is divisible by p40 = 5791987309280956264759051605269253758123 David
                Message 7 of 16 , Dec 24, 2009
                  --- In primeform@yahoogroups.com,
                  Jaroslaw Wroblewski <Jaroslaw.Wroblewski@...> wrote:

                  > My site
                  > http://www.math.uni.wroc.pl/~jwr/cons-fac/
                  > is updated.

                  You might add that N-1 is divisible by
                  p40 = 5791987309280956264759051605269253758123

                  David
                • Jaroslaw Wroblewski
                  2009/12/24 djbroadhurst ... Done. That leaves c221 to factor. Jarek [Non-text portions of this message have been removed]
                  Message 8 of 16 , Dec 24, 2009
                    2009/12/24 djbroadhurst <d.broadhurst@...>

                    > You might add that N-1 is divisible by
                    > p40 = 5791987309280956264759051605269253758123
                    >

                    Done. That leaves c221 to factor.

                    Jarek


                    [Non-text portions of this message have been removed]
                  • Norman Luhn
                    Happy X-Mas to you , primefans ! best wishes Norman __________________________________________________ Do You Yahoo!? Sie sind Spam leid? Yahoo! Mail verfügt
                    Message 9 of 16 , Dec 24, 2009
                      Happy X-Mas to you , primefans !












                      best wishes

                      Norman


                      __________________________________________________
                      Do You Yahoo!?
                      Sie sind Spam leid? Yahoo! Mail verfügt über einen herausragenden Schutz gegen Massenmails.
                      http://mail.yahoo.com

                      [Non-text portions of this message have been removed]
                    • Joe Crump
                      Thanks, you too !!! - Joe Crump ... From: Norman Luhn To: primeform@yahoogroups.com Sent: Thursday, December 24, 2009 1:55 PM Subject: [primeform] 2,3,5,7,....
                      Message 10 of 16 , Dec 24, 2009
                        Thanks, you too !!!

                        - Joe Crump

                        ----- Original Message -----
                        From: Norman Luhn
                        To: primeform@yahoogroups.com
                        Sent: Thursday, December 24, 2009 1:55 PM
                        Subject: [primeform] 2,3,5,7,....



                        Happy X-Mas to you , primefans !

                        best wishes

                        Norman

                        __________________________________________________
                        Do You Yahoo!?
                        Sie sind Spam leid? Yahoo! Mail verfügt über einen herausragenden Schutz gegen Massenmails.
                        http://mail.yahoo.com

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





                        [Non-text portions of this message have been removed]
                      • djbroadhurst
                        Factorization of 18 consecutive 303-digit integers Let x = 9*(2^123 + 63841119); N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216; then for k =
                        Message 11 of 16 , Dec 26, 2009
                          Factorization of 18 consecutive 303-digit integers

                          Let x = 9*(2^123 + 63841119);
                          N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216;
                          then for k = 0 to 17 the factorization of N+k is in
                          http://physics.open.ac.uk/~dbroadhu/cert/ifac18.zip
                          I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
                          It is unlikely that either of the composite integers
                          (N-1)/(3^2*7660729*75495982961295157)
                          (N+18)/(2*811*2777*562633*398311899246763)
                          is divisible by a prime less than 10^35.

                          David Broadhurst, 27 December, 2009
                        • Jaroslaw Wroblewski
                          Congratulations!!! http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Can t wait to see if a 20 is possible to find :-))) Jarek 2009/12/27 djbroadhurst
                          Message 12 of 16 , Dec 26, 2009
                            Congratulations!!!

                            http://www.math.uni.wroc.pl/~jwr/cons-fac/
                            is updated.

                            Can't wait to see if a 20 is possible to find :-)))

                            Jarek

                            2009/12/27 djbroadhurst <d.broadhurst@...>

                            >
                            >
                            > Factorization of 18 consecutive 303-digit integers
                            >
                            > Let x = 9*(2^123 + 63841119);
                            > N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216;
                            > then for k = 0 to 17 the factorization of N+k is in
                            > http://physics.open.ac.uk/~dbroadhu/cert/ifac18.zip<http://physics.open.ac.uk/%7Edbroadhu/cert/ifac18.zip>
                            > I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
                            > It is unlikely that either of the composite integers
                            > (N-1)/(3^2*7660729*75495982961295157)
                            > (N+18)/(2*811*2777*562633*398311899246763)
                            > is divisible by a prime less than 10^35.
                            >
                            > David Broadhurst, 27 December, 2009
                            >
                            >
                            >


                            [Non-text portions of this message have been removed]
                          • djbroadhurst
                            Factorization of 20 consecutive 303-digit integers Let x = 9*(2^123 + 440959466) N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1 then for k
                            Message 13 of 16 , Jan 4, 2010
                              Factorization of 20 consecutive 303-digit integers

                              Let x = 9*(2^123 + 440959466)
                              N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
                              then for k = 0 to 19 the factorization of N+k is in
                              http://physics.open.ac.uk/~dbroadhu/cert/ifac20.zip
                              I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
                              It is unlikely that either of the composite integers
                              (N-1)/(2^2*23*41*290399)
                              (N+20)/(5*19*205889393657*301133059382193691*57657649694534487023)
                              is divisible by a prime less than 10^40.

                              David Broadhurst, 5 January, 2010
                            • Jens Kruse Andersen
                              ... Congratulations on reaching this mark! -- Jens Kruse Andersen
                              Message 14 of 16 , Jan 4, 2010
                                David Broadhurst wrote:
                                > Let x = 9*(2^123 + 440959466)
                                > N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
                                > then for k = 0 to 19 the factorization of N+k is in
                                > http://physics.open.ac.uk/~dbroadhu/cert/ifac20.zip

                                30 November 2009 Jaroslaw Wroblewski wrote:
                                > I estimate that at 300 digits the longest submissions
                                > could be around 20 numbers.

                                Congratulations on reaching this mark!

                                --
                                Jens Kruse Andersen
                              • djbroadhurst
                                ... It was quite tough: only 9/20 came for free; getting the other 11 from OpenPFGW and GMP-ECM was not easy. In fact, I did not find a run of 19 before this
                                Message 15 of 16 , Jan 4, 2010
                                  --- In primeform@yahoogroups.com,
                                  "Jens Kruse Andersen" <jens.k.a@...> wrote:

                                  > Congratulations on reaching this mark!

                                  It was quite tough: only 9/20 came for free; getting the
                                  other 11 from OpenPFGW and GMP-ECM was not easy. In fact,
                                  I did not find a run of 19 before this run of 20, which
                                  came from plugging a hole by extracting
                                  p47 = 21736936592318538144041717556296066161071377743
                                  from
                                  (N+2)/(43*7537*23671) = c293 = p47*p247

                                  David
                                • Jaroslaw Wroblewski
                                  BIG CONGRATULATIONS !!! I am impressed !!! My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Jarek 2010/1/5 djbroadhurst
                                  Message 16 of 16 , Jan 4, 2010
                                    BIG CONGRATULATIONS !!!

                                    I am impressed !!!

                                    My site
                                    http://www.math.uni.wroc.pl/~jwr/cons-fac/
                                    is updated.

                                    Jarek

                                    2010/1/5 djbroadhurst <d.broadhurst@...>

                                    >
                                    >
                                    >
                                    >
                                    > Factorization of 20 consecutive 303-digit integers
                                    >
                                    > Let x = 9*(2^123 + 440959466)
                                    > N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
                                    > then for k = 0 to 19 the factorization of N+k is in
                                    > http://physics.open.ac.uk/~dbroadhu/cert/ifac20.zip<http://physics.open.ac.uk/%7Edbroadhu/cert/ifac20.zip>
                                    > I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
                                    > It is unlikely that either of the composite integers
                                    > (N-1)/(2^2*23*41*290399)
                                    > (N+20)/(5*19*205889393657*301133059382193691*57657649694534487023)
                                    > is divisible by a prime less than 10^40.
                                    >
                                    > David Broadhurst, 5 January, 2010
                                    >
                                    >
                                    >


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