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Re: Yet another factoring puzzle

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  • 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 1 of 23 , Sep 4, 2013
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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 2 of 23 , Sep 16, 2013
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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 3 of 23 , Sep 16, 2013
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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 4 of 23 , Sep 16, 2013
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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 5 of 23 , Sep 18, 2013
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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 6 of 23 , Sep 19, 2013
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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 7 of 23 , Sep 25, 2013
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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
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                  [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 8 of 23 , Sep 27, 2013
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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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