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Re: Yet another factoring puzzle

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  • djbroadhurst
    ... F(n)=((5^n-9)/4)^2-5; if(znorder(Mod(5,29))==14&&Mod(F(9),29)==2,print( proven )); proven David
    Message 1 of 23 , Aug 26, 2013
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      --- In primenumbers@yahoogroups.com,
      Chroma <chromatella@...> wrote:

      > For even k>0 F(14k-5)-2 are divisible by 29
      > How can you prove this relationship?.

      F(n)=((5^n-9)/4)^2-5;
      if(znorder(Mod(5,29))==14&&Mod(F(9),29)==2,print(" proven"));

      proven

      David
    • bobb7772004
      Aurelius surely did NOT mean this to be a square: > 16*F(n) = 5^(2*n)-18*5^n+1= > (5^n-2*5^[(n+1)/2]+1) * (5^n-2*5^[(n+1)/2]+1) --- In
      Message 2 of 23 , Sep 1 1:02 AM
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        Aurelius surely did NOT mean this to be a square: > 16*F(n) = 5^(2*n)-18*5^n+1= > (5^n-2*5^[(n+1)/2]+1) * (5^n-2*5^[(n+1)/2]+1) --- In primenumbers@yahoogroups.com, <d.broadhurst@...> wrote: --- In primenumbers@yahoogroups.com ,
        "mikeoakes2" <mikeoakes2@...> wrote:

        > Aurelius [or somebody] he say:
        >
        > 16*F(n) = 5^(2*n)-18*5^n+1=
        > (5^n-2*5^[(n+1)/2]+1) * (5^n-2*5^[(n+1)/2]+1)

        Wiki, it say:

        > There is evidence dating this algorithm as far back as
        > the Ur III dynasty.

        David
        The second factor should be 5^n PLUS 2*5 etc. Another schoolboy howler.
      • djbroadhurst
        ... Obviously. ... No. Just someome typing good maths even faster than his agile brain was working. David
        Message 3 of 23 , Sep 1 3:44 AM
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          --- In primenumbers@yahoogroups.com, <bobb777@...> wrote:

          > > (5^n-2*5^[(n+1)/2]+1) * (5^n-2*5^[(n+1)/2]+1)
          > The second factor should be 5^n PLUS 2*5 etc.

          Obviously.

          > Another schoolboy howler.

          No. Just someome typing good maths even faster
          than his agile brain was working.

          David
        • j_chrtn
          Hi David, ... F(265) factorization : 2^2 239 739 3001 65482831 256219297361 1693591423953203161 30357718855490049445651878883131113101
          Message 4 of 23 , Sep 1 5:04 AM
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            Hi David,

            --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
            >
            >
            > Exercise 9: Factorize F(265) completely.
            >

            F(265) factorization :
            2^2
            239
            739
            3001
            65482831
            256219297361
            1693591423953203161
            30357718855490049445651878883131113101
            102797863053051886296311988529254933989146480741
            3877828174415305750004694470712933786753853898797149341
            269073781314228860816753896980501633755080784585248485757620731
            9064688734866670980979961151235275322157501870878158907210451074982503913520844038924855331866069982604479273540797473039

            F(263): one composite factor remaining (in progress).

            JL
          • djbroadhurst
            ... Congrats. When I did it, I was running ECM and SNFS in parallel and pulled the plug on SNFS when ECM found a p48. ... Here I used GNFS on the C130. David
            Message 5 of 23 , Sep 1 8:33 AM
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              --- In primenumbers@yahoogroups.com,
              "j_chrtn" <j_chrtn@...> wrote:

              > > Exercise 9: Factorize F(265) completely.
              > F(265) factorization :
              > 2^2
              > 239
              > 739
              > 3001
              > 65482831
              > 256219297361
              > 1693591423953203161
              > 30357718855490049445651878883131113101
              > 102797863053051886296311988529254933989146480741
              > 3877828174415305750004694470712933786753853898797149341
              > 269073781314228860816753896980501633755080784585248485757620731
              > 9064688734866670980979961151235275322157501870878158907210451074982503913520844038924855331866069982604479273540797473039

              Congrats. When I did it, I was running ECM and SNFS in parallel
              and pulled the plug on SNFS when ECM found a p48.

              > F(263): one composite factor remaining (in progress).

              Here I used GNFS on the C130.

              David
            • j_chrtn
              ... I first removed small factors with factorize program from libgmp demos (= trial divisions + Pollard s rho). Then I found all other factors with ECM (using
              Message 6 of 23 , Sep 1 2:06 PM
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                --- In primenumbers@yahoogroups.com, "djbroadhurst" <david.broadhurst@...> wrote:
                >
                >
                > Congrats. When I did it, I was running ECM and SNFS in parallel
                > and pulled the plug on SNFS when ECM found a p48.
                >
                > > F(263): one composite factor remaining (in progress).
                >
                > Here I used GNFS on the C130.
                >
                > David
                >

                I first removed small factors with factorize program from libgmp demos (= trial divisions + Pollard's rho).
                Then I found all other factors with ECM (using P-1 mode for some of them) except for C118 = 3877828174415305750004694470712933786753853898797149341 * 269073781314228860816753896980501633755080784585248485757620731 for which I directly chose cado-nfs.

                For the remaining factor C130 of F(263), I've chosen both ECM and cado-nfs. Still waiting for the result...

                cado-nfs is really a cadeau. Thanks to its authors.

                JL
              • j_chrtn
                ... C130 factorization completed this morning (cado-nfs winner vs ecm). Finally : F(263) = 2^2 11 941 88079 40541279 53849801 3709997079374701
                Message 7 of 23 , Sep 4 11:15 AM
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                  --- In primenumbers@yahoogroups.com, "djbroadhurst" <david.broadhurst@...> wrote:
                  >
                  >
                  > > F(263): one composite factor remaining (in progress).
                  >
                  > Here I used GNFS on the C130.
                  >
                  > David
                  >

                  C130 factorization completed this morning (cado-nfs winner vs ecm).

                  Finally :
                  F(263) =
                  2^2
                  11
                  941
                  88079
                  40541279
                  53849801
                  3709997079374701
                  23529341871144986702279
                  7270487490315018281073601513510602536818804246566820732218199
                  790942341954447264420872400154902667291367695120485038995898872524619
                  711892421814353474455471503465724397364909744377767780766071778400352308618205366660863738451363497318680099295967052261874590114183928845236941340734473602381665606275060651

                  JL
                • djbroadhurst
                  ... Congrats, again, J-L. This link should show all the factorizations from n=261 to n=290: http://preview.tinyurl.com/mx3cdoe David
                  Message 8 of 23 , Sep 4 1:02 PM
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                    --- In primenumbers@yahoogroups.com,
                    "j_chrtn" <j_chrtn@...> wrote:

                    > F(263) =

                    Congrats, again, J-L.
                    This link should show all the factorizations from n=261 to n=290:
                    http://preview.tinyurl.com/mx3cdoe

                    David
                  • djbroadhurst
                    ... As you can see, Exercise 8 was censored :-) As far as I can tell, no-one (apart from the setter) yet solved Exercise 6, which can be done in less than 2
                    Message 9 of 23 , Sep 16 12:10 PM
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                      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

                      > Definition: Let F(n) = ((5^n-9)/4)^2-5 for integer n > 0.
                      >
                      > Exercise 1: For even n > 2, prove that F(n) is composite.
                      >
                      > Exercise 2: For odd n > 1, prove that F(n)/4 is composite.
                      >
                      > Exercise 3: For k > 1, prove that F(3*k) has at least 4 odd
                      > prime divisors.
                      >
                      > Exercise 4: Factorize F(6) = 15241211 completely, by hand.
                      >
                      > Exercise 5: Find the complete factorization of F(n) for at
                      > least one odd integer n > 250.
                      >
                      > Exercise 6: Find the complete factorization of F(n) for at
                      > least one even integer n > 600.
                      >
                      > Exercise 7: Factorize F(263) completely.
                      >
                      > Exercise 9: Factorize F(265) completely.

                      As you can see, Exercise 8 was censored :-)

                      As far as I can tell, no-one (apart from the setter)
                      yet solved Exercise 6, which can be done in less
                      than 2 minutes, using OpenPFGW. What is remarkable
                      about this exercise is that it can be solved so
                      quickly. Heuristically, that was not to be expected.

                      Thanks to Bernardo Boncompagni, Mike Oakes and
                      Jean-Louis Charton, for disposing neatly of the
                      other exercises, and to Ben Buhrow, for his
                      excellent package, Yafu.

                      David
                    • j_chrtn
                      ... For sure one can find this solution quickly using openpfgw. But the smart guy now knowing that you (or others) have filled in factordb with many numbers of
                      Message 10 of 23 , Sep 16 4:00 PM
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                        --- In primenumbers@yahoogroups.com, <david.broadhurst@...> wrote:

                        --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:


                        > As you can see, Exercise 8 was censored :-)
                        >
                        > As far as I can tell, no-one (apart from the setter)
                        > yet solved Exercise 6, which can be done in less
                        > than 2 minutes, using OpenPFGW. What is remarkable
                        > about this exercise is that it can be solved so
                        > quickly. Heuristically, that was not to be expected.

                        For sure one can find this solution quickly using openpfgw.

                        But the smart guy now knowing that you (or others) have filled in factordb with many numbers of this form just has to type http://factordb.com/index.php?query=%28%285^n-9%29%2F4%29^2-5&use=n&perpage=20&format=1&sent=1&PR=1&PRP=1&C=1&CF=1&U=1&FF=1&VP=1&EV=1&OD=1&VC=1&n=600

                        to find the solution.

                         

                        :-)

                         

                        J-L 

                         

                        PS: I really did F(263) and F(265) factorization completely. I did not check factordb.

                         

                      • djbroadhurst
                        ... Indeed, it was I who added http://factordb.com/index.php?id=1100000000464478896 with factors p404 and p414. Jean-Louis gets full marks for exploiting
                        Message 11 of 23 , Sep 16 5:08 PM
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                          Jean-Louis Charton wrote:

                          > For sure one can find this solution quickly using openpfgw.
                          > But the smart guy now knowing that you (or others) have filled
                          > in factordb with many numbers of this form just has to type...

                          Indeed, it was I who added
                          http://factordb.com/index.php?id=1100000000464478896
                          with factors p404 and p414.

                          Jean-Louis gets full marks for exploiting factordb,
                          instead of doing what I intended, along the lines of

                          $ more abc

                          ABC2 25^$a-4*5^$a-1&25^$a+4*5^$a-1
                          a: from 301 to 304

                          $ pfgw -f -d -e20000000 abc

                          25^301-4*5^301-1 has factors: 2^2*7229
                          25^302-4*5^302-1 has factors: 2^2*3173791
                          25^303-4*5^303-1 has factors: 2^2*1931*934579
                          25^304-4*5^304-1 has factors: 2^2*11*29*1289*1759*9511*27851
                          (25^304-4*5^304-1)/(2^2*11*29*1289*1759*9511*27851) is 3-PRP!
                          (0.0057s+0.2797s)
                          25^304+4*5^304-1 has factors: 2^2*1439*17390951
                          (25^304+4*5^304-1)/(2^2*1439*17390951) is 3-PRP!
                          (0.0058s+0.3055s)

                          But now, J-L, are you able to explain my comment:

                          > What is remarkable about this exercise is that it can be
                          > solved so quickly. Heuristically, that was not to be expected.

                          Can you quantify my suprise?

                          AmitiƩs

                          David
                        • j_chrtn
                          ... Well, I would say that for n = 600, both algebraic factors are more than 415 digits and trial factoring them with pfgw up to 20000000 with -d option you
                          Message 12 of 23 , Sep 18 4:39 PM
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                            --- In primenumbers@yahoogroups.com, <david.broadhurst@...> wrote:

                            >But now, J-L, are you able to explain my comment:
                            >
                            >> What is remarkable about this exercise is that it can be
                            >> solved so quickly. Heuristically, that was not to be expected.
                            >
                            >Can you quantify my suprise?
                            >

                            Well, I would say that for n >= 600, both algebraic factors are more than 415 digits and trial factoring them with pfgw up to 20000000 with -d option you may remove a 20000000-smooth composite factor of say 10 or 15 digits. This leave 2 factors of more than 400 digits. The "probability" for one such factor to be prime is roughly 1/l(10^400), that is about 0.001. So for both factors to be prime the probability is about 0.000001.

                            Am I right ?

                            Amicalement,

                            J-L
                          • djbroadhurst
                            ... We seek complete factorization of both of the cofactors 25^k-4*5^k-1 and 25^k+4*5^k-1 for some k 300, where each has more than 400 decimal digits. We had
                            Message 13 of 23 , Sep 19 12:26 PM
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                              Jean-Louis Carton wrote:

                              > So for both factors to be prime the probability is about 0.000001.

                              We seek complete factorization of both of the cofactors
                              25^k-4*5^k-1 and 25^k+4*5^k-1
                              for some k > 300, where each has more than
                              400 decimal digits. We had better avoid
                              the case k = 0 mod 3, where each cofactor
                              has an algebraic factorization, which is
                              here a distinct disadvantage.

                              Suppose that we sieve out primes to depth d
                              and hope for what is left to yield a
                              a pair of PRPs as here, with k = 304:

                              (25^304-4*5^304-1)/(2^2*11*29*1289*1759*9511*27851) is 3-PRP!
                              (25^304+4*5^304-1)/(2^2*1439*17390951) is 3-PRP!

                              The probability for success for a single value of k
                              coprime to 3 is of order

                              (exp(Euler)*log(p)/log(25))^2/k^2

                              Setting p = 2*10^7 and summing over /all/ k > 300,
                              the probability that Exercise 6 has /no/ solution is

                              exp(-2/3*(exp(Euler)*log(2*10^7)/log(25))^2/301) =~ 83%

                              In fact it has a solution almost immediately, at k = 304.

                              David
                            • Phil Carmody
                              ... But is that more or less remarkable than the expectation of any one of Phil Taylor s darts landing in the region around where it actually landed? You only
                              Message 14 of 23 , Sep 25 3:08 PM
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                                On Mon, 9/16/13, djbroadhurst wrote:
                                --- "djbroadhurst" <d.broadhurst@...> wrote:
                                > > Exercise 6: Find the complete factorization of F(n) for at
                                > > least one even integer n > 600.
                                > As far as I can tell, no-one (apart from the setter)
                                > yet solved Exercise 6, which can be done in less
                                > than 2 minutes, using OpenPFGW. What is remarkable
                                > about this exercise is that it can be solved so
                                > quickly. Heuristically, that was not to be expected.

                                But is that more or less remarkable than the expectation of
                                any one of Phil Taylor's darts landing in the region around
                                where it actually landed? You only chose that target after
                                the arrow had landed, I'm sure.

                                How many mathematical diversions have you looked at
                                via the medium of numerical computation? How many of
                                them would you expect to be remarkably easier than
                                expected to solve? Probably a non-zero answer. Don't
                                be surprised that one particular example was one.

                                Knowing what he's trying to say, even if he's not getting
                                it across clearly,
                                Phil
                                --
                                () ASCII ribbon campaign () Hopeless ribbon campaign
                                /\ against HTML mail /\ against gratuitous bloodshed

                                [stolen with permission from Daniel B. Cristofani]
                              • djbroadhurst
                                ... It happened thus: 1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for n
                                Message 15 of 23 , Sep 27 2:52 PM
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                                   ---In primenumbers@yahoogroups.com, <thefatphil@...> wrote:

                                   

                                  > You only chose that target after
                                  > the arrow had landed, I'm sure.

                                   

                                  It happened thus:

                                   

                                  1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for
                                  n <= 300, completely. As later shown in "factordb", I succeeded.

                                   

                                  2) Meanwhile I ran OpenPFGW on n in [301,600], hoping for a
                                  quick outlier and found none.

                                   

                                  3) I estimated the probability of an easily discoverable
                                  completely factorization for n>600 and found it to be small.

                                   

                                  4) Recalling how I had once been caught out before by
                                  a "probably no more" heuristic, I set a lone process running on
                                  n in [601, 10000]  so as not to be caught out again by Jens.

                                   

                                  5) When I later looked  and pfgw.log, it had found a hit at
                                  n=608.

                                   

                                  So yes, Phil, you are quite correct that the puzzle was set
                                  after this finding. However the heuristic that I gave was
                                  made prior to my discovery, else I would not have said that
                                  I was surprised.

                                   

                                  The point that you are making (I think) is that I do such 
                                  expsriments often and only notice when the result is unexpected.
                                  I don't tell folk about all the boring times when a negative
                                  heuristic is borne out by a null result. That is the selection

                                  effect.

                                   

                                  David (guilty of not boring folk with what is routine)

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