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Primes in the form 2^n+3^n

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  • Di Maria Giovanni
    What about the form 2^n+3^n ? Do primes exist in this form? There are sites about this form? Thank you Regards Giovanni Di Maria
    Message 1 of 12 , Oct 22, 2010
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      What about the form 2^n+3^n ?
      Do primes exist in this form?
      There are sites about this form?
      Thank you
      Regards
      Giovanni Di Maria
    • Peter Kosinar
      Hi Giovanni, ... 2^0 + 3^0 = 2 is a prime! 2^1 + 3^1 = 5 is a prime! 2^2 + 3^2 = 13 is a prime! ... proof that it works for all values of n is left as an
      Message 2 of 12 , Oct 22, 2010
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        Hi Giovanni,

        > What about the form 2^n+3^n ?
        > Do primes exist in this form?

        2^0 + 3^0 = 2 is a prime!
        2^1 + 3^1 = 5 is a prime!
        2^2 + 3^2 = 13 is a prime!
        ... proof that it works for all values of 'n' is left as an exercise for
        the reader. :-)

        Now, on a more serious note. The only other value of n which produces a
        prime and which I know is n=4 (2^4 + 3^4 = 97). Other than that, this
        form is going to be very very rare -- since if X is an odd number,
        (a^X + b^X) = (a+b)(a^(X-1).b^0 - a^(X-2).b^1 + ... +/- a^0.b^(X-1)).

        Thus, if 'n' has any odd factor, the result is a composite. Put slightly
        differently, it'd better be a power of two if the result is to be prime.
        Fermat primes would be proud of their cousins. :-)

        Peter
      • Phil Carmody
        ... They are one form of generalised fermat primes. They have algebraic factors if n has any odd factors, and so can only be prime if n is a power of 2. (Just
        Message 3 of 12 , Oct 22, 2010
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          > What about the form 2^n+3^n ?
          > Do primes exist in this form?
          > There are sites about this form?

          They are one form of generalised fermat primes. They have algebraic factors if n has any odd factors, and so can only be prime if n is a power of 2. (Just like Fermat Numbers.)

          Phil*
        • Jack Brennen
          Is there any place online that collects factorizations of numbers of the form a^(2^x)+b^(2^x) ? Just wondering; I just spent a few CPU minutes finding the
          Message 4 of 12 , Oct 22, 2010
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            Is there any place online that collects factorizations of numbers of
            the form a^(2^x)+b^(2^x) ?

            Just wondering; I just spent a few CPU minutes finding the complete
            factorization of the 123-digit number:

            2^256 + 3^256 ==
            72222721 *
            343200070657 *
            2226198380033 *
            3376663028737 *
            1839605176202823817996787333633 *
            405549420455750246193993361998354279613273199617

            and it seems like one of those things that people like to put into
            online databases...

            On 10/22/2010 12:45 PM, Phil Carmody wrote:
            >> What about the form 2^n+3^n ?
            >> Do primes exist in this form?
            >> There are sites about this form?
            >
            > They are one form of generalised fermat primes. They have algebraic factors if n has any odd factors, and so can only be prime if n is a power of 2. (Just like Fermat Numbers.)
            >
            > Phil*
            >
            >
            >
            >
            >
            > ------------------------------------
            >
            > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
            > The Prime Pages : http://www.primepages.org/
            >
            > Yahoo! Groups Links
            >
            >
            >
            >
            >
          • djbroadhurst
            ... That was already in http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt which also reveals that 2^(2^9) + 3^(2^9) == 4043777 * 57987375533057 *
            Message 5 of 12 , Oct 22, 2010
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              --- In primenumbers@yahoogroups.com,
              Jack Brennen <jfb@...> wrote:

              > 2^256 + 3^256 ==
              > 72222721 *
              > 343200070657 *
              > 2226198380033 *
              > 3376663028737 *
              > 1839605176202823817996787333633 *
              > 405549420455750246193993361998354279613273199617

              That was already in
              http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt
              which also reveals that

              2^(2^9) + 3^(2^9) ==
              4043777 *
              57987375533057 *
              406297848379393 *
              296909426637032277938768757025793 *
              1440276254252769698681054259953238091664449537 *
              2236335363090199099071989377913382509011409921 *
              212085278921429005793934248630969041682071162298569605814071345111198398726196355614721

              David
            • Jens Kruse Andersen
              ... You can Google numbers. http://www.google.com/search?q=72222721+prime gives http://www.research.att.com/~njas/sequences/A094499 which links
              Message 6 of 12 , Oct 22, 2010
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                Jack Brennen wrote:
                > Is there any place online that collects factorizations of numbers of
                > the form a^(2^x)+b^(2^x) ?

                You can Google numbers.
                http://www.google.com/search?q=72222721+prime gives
                http://www.research.att.com/~njas/sequences/A094499 which links
                http://www.ams.org/journals/mcom/1998-67-221/S0025-5718-98-00891-6/S0025-5718-98-00891-6.pdf
                The paper is from 1998. See page 8.

                --
                Jens Kruse Andersen
              • julienbenney
                I had a look on the factorisation site http://www.alpertron.com.ar/ECM.HTM and it easily found the smallest factor of the 489-digit number 2^512+3^512.
                Message 7 of 12 , Oct 24, 2010
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                  I had a look on the factorisation site "http://www.alpertron.com.ar/ECM.HTM" and it easily found the smallest factor of the 489-digit number 2^512+3^512. However, so far I have not found a further factor of the 477-digit cofactor, although it certainly is not prime.

                  I know well that F9(10) or 10^512+1 is not yet completely factored (in fact, I once tried to spend several nights on my mediocre home computer trying to find more than the three known factors) so it would seem reasonable that 2^(2^10)+3^(2^10) would also have not been fully factored. Is any other prime factor of 2^(2^10)+3^(2^10) known?
                • Paul Leyland
                  ... In http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt you will find the complete factorization:
                  Message 8 of 12 , Oct 25, 2010
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                    > I had a look on the factorisation site "http://www.alpertron.com.ar/ECM.HTM"
                    > and it easily found the smallest factor of the 489-digit number 2^512+3^512.
                    > However, so far I have not found a further factor of the 477-digit cofactor,
                    > although it certainly is not prime.

                    In http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt you
                    will find the complete factorization:

                    4043777.57987375533057.406297848379393.296909426637032277938768757025793.1440276254252769698681054259953238091664449537.2236335363090199099071989377913382509011409921. P87

                    Paul
                  • djbroadhurst
                    ... The original poster seemed to write 512 where he means 2^10. It is (2^1024+3^1024)/2330249132033 that has 477 digits and is not factorized at Paul s fine
                    Message 9 of 12 , Oct 25, 2010
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                      --- In primenumbers@yahoogroups.com,
                      Paul Leyland <paul@...> wrote:

                      > > so far I have not found a further factor
                      > > of the 477-digit cofactor
                      > In http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt you
                      > will find the complete factorization:

                      The original poster seemed to write 512 where he means 2^10.

                      It is (2^1024+3^1024)/2330249132033 that has 477 digits
                      and is not factorized at Paul's fine site.

                      David
                    • Paul Leyland
                      ... Ah. Ask and ye shall receive, but first make sure you know what you are asking for! I m running a few curves on 2^1024+3^1024 and will post again if
                      Message 10 of 12 , Oct 25, 2010
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                        On Mon, 2010-10-25 at 09:57 +0000, djbroadhurst wrote:
                        >
                        > --- In primenumbers@yahoogroups.com,
                        > Paul Leyland <paul@...> wrote:
                        >
                        > > > so far I have not found a further factor
                        > > > of the 477-digit cofactor
                        > > In http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt you
                        > > will find the complete factorization:
                        >
                        > The original poster seemed to write 512 where he means 2^10.
                        >
                        > It is (2^1024+3^1024)/2330249132033 that has 477 digits
                        > and is not factorized at Paul's fine site.
                        >
                        > David

                        Ah. Ask and ye shall receive, but first make sure you know what you are
                        asking for!

                        I'm running a few curves on 2^1024+3^1024 and will post again if
                        anything turns up.

                        Paul
                      • Robert Gerbicz
                        2010/10/25 djbroadhurst ... http://www1.uni-hamburg.de/RRZ/W.Keller/GFNsmall.html gives one more prime factor:
                        Message 11 of 12 , Oct 25, 2010
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                          2010/10/25 djbroadhurst <d.broadhurst@...>

                          >
                          >
                          >
                          >
                          > --- In primenumbers@yahoogroups.com <primenumbers%40yahoogroups.com>,
                          > Paul Leyland <paul@...> wrote:
                          >
                          > > > so far I have not found a further factor
                          > > > of the 477-digit cofactor
                          > > In http://www.leyland.vispa.com/numth/factorization/anbn/3+2.txt you
                          > > will find the complete factorization:
                          >
                          > The original poster seemed to write 512 where he means 2^10.
                          >
                          > It is (2^1024+3^1024)/2330249132033 that has 477 digits
                          > and is not factorized at Paul's fine site.
                          >
                          > David
                          >
                          >
                          >
                          http://www1.uni-hamburg.de/RRZ/W.Keller/GFNsmall.html gives one more prime
                          factor: 10176954088500686156890644481 and remains c449.


                          [Non-text portions of this message have been removed]
                        • Paul Leyland
                          ... Found both of those now: [2010-10-25 10:16:35 GMT] n10: probable factor returned by paul@leyland.vispa.com (mesh_4:v2.0k)! Factor=2330249132033
                          Message 12 of 12 , Oct 25, 2010
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                            > > The original poster seemed to write 512 where he means 2^10.
                            > >
                            > > It is (2^1024+3^1024)/2330249132033 that has 477 digits
                            > > and is not factorized at Paul's fine site.

                            > http://www1.uni-hamburg.de/RRZ/W.Keller/GFNsmall.html gives one more prime
                            > factor: 10176954088500686156890644481 and remains c449.

                            Found both of those now:

                            [2010-10-25 10:16:35 GMT] n10: probable factor returned by
                            paul@... (mesh_4:v2.0k)! Factor=2330249132033 Method=ECM
                            B1=250000 Sigma=90206226
                            [2010-10-25 10:16:35 GMT] n10: Composite factor returned by
                            paul@...!
                            Factor=160236879228127195164826503514143273848644887617419237583589880672819224787962378613168358191054320882014589878763655598739442603012525981141289393946718413503071756942365084833216516319583697404684336969516432404909365911461501691450758224189677519380661024336610343945039247522150954814862123259434023865815579858133945351151726697231584694926774191784591080286025985084327594874779310399776681428659805398074761131932929563211305463464053349254128040769205414113291414263809 Method=ECM B1=250000 Sigma=90206226


                            [2010-10-25 12:33:42 GMT] n10_1: probable factor returned by pcl@anubis
                            (anubis5)! Factor=10176954088500686156890644481 Method=ECM B1=250000
                            Sigma=1927001934
                            [2010-10-25 12:33:42 GMT] n10_1: Composite factor returned by
                            pcl@anubis!
                            Factor=15745072428810966584413689569028050952121723170588717492356803041962045543719421466634314626515681367806038954950649466918177284600127897807223229616734493963870994176961152031838265180914651227518129736337432709373581447583480385668299052789615157720829985754126147818607443036795542666041266727368967991929574179247057328372407884419914237683940826611054069893545018041498309020902084444177674786383308058081865530163424306601965434254433964505089 Method=ECM B1=250000 Sigma=1927001934


                            I'll leave it in the ECMserver for the moment as it is being given a
                            relatively small fraction of my resources.


                            Paul
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