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Re: Two large consecutive smooth numbers

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  • Kermit Rose
    ... e = limit(n-- infinity) (1+1/n)^n ln((n+1)/n) = ln(1 + 1/n) ln( (1+1/n)^n) is, for large n, approximately equal to 1. ln((1+1/n)^n) = n ln(1+1/n)
    Message 1 of 17 , Mar 4, 2012
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      On 3/4/2012 7:59 AM, primenumbers@yahoogroups.com wrote:
      > 1a. Two large consecutive smooth numbers
      > Posted by: "WarrenS"warren.wds@... warren_d_smith31
      > Date: Sat Mar 3, 2012 9:19 am ((PST))
      >
      > N=43623575184339996059537425773119366447006380455838\
      > 696504055889999185302903791148393125043181272726633463298672436846034128
      >
      > and N+1, both are "smooth", i.e both factor entirely into primes<=9168769.
      >
      > Can you do better (i.e. make N larger, and the max prime smaller)?
      >

      e = limit(n--> infinity) (1+1/n)^n


      ln((n+1)/n) = ln(1 + 1/n)

      ln( (1+1/n)^n) is, for large n, approximately equal to 1.


      ln((1+1/n)^n) = n ln(1+1/n)

      ln(1+1/n) = approximately (1/n) for large n.


      Find solutions to k1 ln(2) + k2 ln(3) + k3 ln(5) + .... < 1/n for
      target large values of n.


      Minimize k1 ln(2) + k2 ln(3) + k3 ln(5) + .... where some of the k's
      are required to be positive, and some negative.


      k1 ln(2) + k2 ln(3) is approximately 0

      k1 ln(2) = approximately - k2 ln(3)

      -k1/k2 = approximately ln(3)/ln(2)

      One of k1, k2 is positive. The other is negative.

      It seems straight forward to calculate the coefficients k1, k2, ,k3, etc
      which minimizes this sum for given sets of primes,
      2,3,5,etc.

      From that minimum value of the sum for given sets of primes,
      calculate n = int(1/minimum) as an upper bound for the n which applies.



      Kermit Rose
    • Kermit Rose
      ... To construct quadratics, x^2 - b x + c and x^2 - b x + c + 1 which are both factorisable, we look for integers b, k1, k2 such that c = k1 * (b-k1) and c+1
      Message 2 of 17 , Mar 4, 2012
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        On 3/4/2012 7:59 AM, primenumbers@yahoogroups.com wrote:
        > 1c. Re: Two large consecutive smooth numbers
        > Posted by: "Phil Carmody"thefatphil@... thefatphil
        > Date: Sat Mar 3, 2012 3:03 pm ((PST))
        >
        > Calling Doctor Broadhurst for suggestion of the best metric by which to evaluate such records. A simple log doesn't necessarily tell the whole tale at all.
        >
        > Be warned, Warren - Dr. B is sitting on a corpus of algebraic formulae such that p(x) and p(x)+1 have algebraic factorisations, which makes smoothness measurably (I was going to say immeasurably, and then realised the stupidity of such a word choice) more likely.
        >
        > I'll not play this game, as I have an appointment with 21 farmers in Lithuania (otherwise known as the biggest brewery crawl yet...)
        >
        > Phil

        To construct quadratics,

        x^2 - b x + c and x^2 - b x + c + 1 which are both factorisable,

        we look for integers b, k1, k2 such that

        c = k1 * (b-k1)
        and
        c+1 = k2 * (b-k2)

        1*3 = 3; 2 * 2 = 4

        x^2 - 4 x + 3 = (x-1)*(x-3)
        x^2 - 4 x + 4 = (x-2)^2

        2 * 4 = 8; 3 * 3 = 9

        x^2 - 6 x + 8 = (x-2)*(x-4)
        x^2 - 6 x + 9 = (x-3)^2 which is really the same as the first example
        translated by 1.

        I will guess that maybe the only quadratic polynomial solutions are
        translations of the first example.

        Cubic polynomial solutions should be more prolific.

        Kermit
      • Jim White
          Hi kermit,   The quadratic case simply corresponds to a 3-tuple of smooth {Q, Q+1, Q+2} inferring a pair at {(Q+1)^2 - 1, (Q+1)^2}   We are aiming for
        Message 3 of 17 , Mar 4, 2012
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          Hi kermit,
           
          The quadratic case simply corresponds to a 3-tuple
          of smooth {Q, Q+1, Q+2} inferring a pair at
          {(Q+1)^2 - 1, (Q+1)^2}
           
          We are aiming for more massive "power boosting".
           
          cf: Prouhet-Tarry-Escott problem.
           
          Warrens SMODA II document has a good list of
          known cases for polynomials to orders up to 10,
          and I believe he has found pairs using a 12-th degree
          poly.
           
          Cheers!


          ________________________________
          From: Kermit Rose <kermit@...>
          To: primenumbers@yahoogroups.com
          Sent: Monday, 5 March 2012, 1:59
          Subject: [PrimeNumbers] Re: Two large consecutive smooth numbers

          > I will guess that maybe the only quadratic polynomial solutions are
          > translations of the first example.

          Cubic polynomial solutions should be more prolific.

          Kermit




          [Non-text portions of this message have been removed]
        • WarrenS
          Jim White & I have been trying to construct these things because they are grist for my new factoring algorithm SMODA. That was one example of our output.
          Message 4 of 17 , Mar 4, 2012
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            Jim White & I have been trying to construct these things because they are grist for my
            new factoring algorithm SMODA. That was one example of our output. Although we can break (and have broken) that record, I could only make N have about 30% more digits before
            my current program would get very slow or self destruct.

            If you have new approaches, and can break that record using them, great.
            The PTE approach Broadhurst & Carmody were hinting at, is what I am using now for the
            largest ones, so that's not a new idea unless you know something I do not about it.

            If you are interested in donating computer time to this effort, intel last 10 years,
            unix variant, ok to run weeks at a time in background, then
            email warren.wds AT gmail.com.

            Thank you.
          • djbroadhurst
            ... Yes. Let {f(y)= (y^2-11^4)*(y^2-35^2)*(y^2-47^2)* (y^2-94^2)*(y^2-146^2)*(y^2-148^2)/ 67440294559676054016000 - 1;} With N = f(1210851834572), we find that
            Message 5 of 17 , Mar 4, 2012
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              --- In primenumbers@yahoogroups.com,
              "WarrenS" <warren.wds@...> wrote:

              > N=43623575184339996059537425773119366447006380455838\
              > 696504055889999185302903791148393125043181272726633463298672436846034128
              > and N+1, both are "smooth", i.e both factor entirely into
              > primes <= 9168769.
              > Can you do better (i.e. make N larger, and the max prime smaller)?

              Yes. Let
              {f(y)=
              (y^2-11^4)*(y^2-35^2)*(y^2-47^2)*
              (y^2-94^2)*(y^2-146^2)*(y^2-148^2)/
              67440294559676054016000 - 1;}

              With N = f(1210851834572),
              we find that N*(N+1) is 5205793-smooth.

              With N = f(1606741747790),
              we find that N*(N+1) is 8686687-smooth.

              David
            • djbroadhurst
              ... I have not been following this in detail, but I gained the impression that Warren s original advert was far too optimistic and that now his heuristic for
              Message 6 of 17 , Mar 4, 2012
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                --- In primenumbers@yahoogroups.com,
                "WarrenS" <warren.wds@...> wrote:

                > Jim White & I have been trying to construct these things
                > because they are grist for my new factoring algorithm SMODA.

                I have not been following this in detail, but I gained the impression
                that Warren's original advert was far too optimistic and that now
                his heuristic for oracular factorization has escalated from
                exp(log(N)^(1/3+o(1))) to the far less encouraging
                exp(log(N)^(2/3+o(1))) as the time to construct a database.

                Is this a fair summary of the setback?

                David
              • WarrenS
                ... --well, yes and no. (And it wasn t a setback since I knew it all along.) (1) My factorization algorithm SMODA as far as I know still works and still
                Message 7 of 17 , Mar 4, 2012
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                  --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
                  >
                  >
                  >
                  > --- In primenumbers@yahoogroups.com,
                  > "WarrenS" <warren.wds@> wrote:
                  >
                  > > Jim White & I have been trying to construct these things
                  > > because they are grist for my new factoring algorithm SMODA.
                  >
                  > I have not been following this in detail, but I gained the impression
                  > that Warren's original advert was far too optimistic and that now
                  > his heuristic for oracular factorization has escalated from
                  > exp(log(N)^(1/3+o(1))) to the far less encouraging
                  > exp(log(N)^(2/3+o(1))) as the time to construct a database.
                  >
                  > Is this a fair summary of the setback?

                  --well, yes and no. (And it wasn't a "setback" since I knew it all along.)

                  (1) My factorization algorithm SMODA as far as I know still works and still runs in
                  exp(log(N)^(1/3+-o(1))) time PROVIDED database ("oracle") is available for its use.
                  It is plausibly better under this proviso than quadratic sieve and number field sieve,
                  but that at present is unconfirmed.

                  (2) For the problem of computing the database, however, I only have
                  exp(log(N)^(2/3+-o(1))) time algorithms for. However as we just saw, the o(1)
                  is fairly beneficial, since we can reach at least 400-bit-long database entries on a single computer, indeed Broadhurst just found some database entries of that size
                  in a matter of a few hours -- pretty fast turnaround! (His weren't as large as my
                  best records, but obviously Broadhurst has already built a search code comparable to or better than mine.) In fact I hope to release a preliminary database by me & Jim White, going up to 400-bits, in a few more days to interested parties.

                  (3) You might say that (2) sort of demolishes (1), but that is debatable. The thing is,
                  the database-build is something that all factorers worldwide can do collaboratively and do only once. Therefore, it is not fair to judge this runtime on the same footing as the other runtime. I admit I'm not quite sure how to judge it, because it has been a fairly rare thing
                  in the world so far, to have oracle-algorithms that actually are useful.

                  It is conceivable that (2)'s theoretical runtime can be sped up, but at present, I haven't been able to. Furthermore, few or no experts have carefully examined either (1) or (2) yet so it remains possible I'm crazy and the whole thing is broken. I doubt that -- I think any remaining errors are minor -- I'm just giving you fair warning.

                  --Warren D Smith
                • Jim White
                  Hard puzzle, really hard puzzle   We know also that, while max N might exist with ~5000 digits, his nearest p-smooth neighbour pair might well be hundreds of
                  Message 8 of 17 , Mar 5, 2012
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                    Hard puzzle, really hard puzzle
                     
                    We know also that, while max N might exist with ~5000
                    digits, his nearest p-smooth neighbour pair might
                    well be hundreds of digits smaller.  
                     
                    What we don't know is which pairs N are "PTE-compatible",
                    ie can be found via some factoring
                    polynomial whose roots are all p-smooth.
                     
                    Any ideas on that issue would be useful
                     
                    Jim White
                     

                    ________________________________
                    From: Andrey Kulsha <Andrey_601@...>
                    To: PrimeNumbers@...
                    Sent: Sunday, 4 March 2012, 9:53
                    Subject: Re: [PrimeNumbers] Two large consecutive smooth numbers


                     

                    Heuristically, log(max_N) is nearly proportional to sqrt(max_prime).

                    So, with p < 9168769, one can find N with more than 5000 digits.

                    But that's a hard puzzle, really.

                    [Non-text portions of this message have been removed]
                  • Jim White
                    Andrey s chain puzzle is interesting.  Could it be he already has found the maximum possible result for chain length 13?   It s hard to see how that result
                    Message 9 of 17 , Mar 5, 2012
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                      Andrey's chain puzzle is interesting.  Could it be
                      he already has found the maximum possible result
                      for chain length 13?
                       
                      It's hard to see how that result can be beaten.
                       
                      Some results with weights of 2.2 or more:
                       
                          28246112570058, weight = 2.2053 (P =  1257251)
                          18911412089528, weight = 2.2077 (P =  1032307)
                         218381019281507, weight = 2.2410 (P =  2504167)
                           9288363679368, weight = 2.2480 (P =   587149)
                        3393509932556102, weight = 2.2536 (P =  7788997)
                        4532039198639948, weight = 2.2536 (P =  8856259)
                        4532039198639949, weight = 2.2536 (P =  8856259)
                       12469670986534198, weight = 2.2547 (P = 13762769)
                       10160468895884110, weight = 2.2592 (P = 12163843)
                         461881571558141, weight = 2.2615 (P =  3050603)
                        7909529450841510, weight = 2.2621 (P = 10669823)
                         211814723372355, weight = 2.2918 (P =  1782043)
                         430753934627814, weight = 2.4217 (P =  1103933)

                      Perhaps the 14-chain at N = 4532039198639948 might
                      be a good result? What are the best known results
                      for 14 or longer chains?


                      ________________________________
                      From: Andrey Kulsha <Andrey_601@...>
                      To: PrimeNumbers@...
                      Sent: Sunday, 4 March 2012, 9:53
                      Subject: Re: [PrimeNumbers] Two large consecutive smooth numbers


                       

                      > Puzzle: find a chain of 13 consecutive p-smooth integers,
                      > starting at N, with log(N)/log(p) greater than
                      >
                      > log(8559986129664)/log(58393) = 2.71328

                      Best regards,

                      Andrey




                      [Non-text portions of this message have been removed]
                    • Andrey Kulsha
                      ... No, I think that log/log ratio has no limit. ... Brute force search yielded: N = 505756884840 for 14-chain N = 285377140980 for 15-chain N = 32290958458
                      Message 10 of 17 , Mar 5, 2012
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                        > Andrey's chain puzzle is interesting. Could it
                        > be he already has found the maximum possible
                        > result for chain length 13?

                        No, I think that log/log ratio has no limit.

                        > Perhaps the 14-chain at N = 4532039198639948
                        > might be a good result? What are the best known
                        > results for 14 or longer chains?

                        Brute force search yielded:
                        N = 505756884840 for 14-chain
                        N = 285377140980 for 15-chain
                        N = 32290958458 for 16-chain
                        as listed in http://www.primefan.ru/stuff/math/maxs.xls
                        (there k+1 is chain length)

                        Best regards,

                        Andrey
                      • Jim White
                        Andrey,   I can t use that file, I don t have XL.  Any chance of a text export? eg comma-separated fields     ________________________________ From:
                        Message 11 of 17 , Mar 5, 2012
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                          Andrey,
                           
                          I can't use that file, I don't have XL.  Any chance
                          of a text export? eg comma-separated fields
                           
                           

                          ________________________________
                          From: Andrey Kulsha <Andrey_601@...>
                          To: PrimeNumbers@...
                          Sent: Tuesday, 6 March 2012, 6:28
                          Subject: [PrimeNumbers] Re: 13-chains of consecutive smooth numbers



                           

                          > Andrey's chain puzzle is interesting. Could it
                          > be he already has found the maximum possible
                          > result for chain length 13?

                          No, I think that log/log ratio has no limit.

                          > Perhaps the 14-chain at N = 4532039198639948
                          > might be a good result? What are the best known
                          > results for 14 or longer chains?

                          Brute force search yielded:
                          N = 505756884840 for 14-chain
                          N = 285377140980 for 15-chain
                          N = 32290958458 for 16-chain
                          as listed in http://www.primefan.ru/stuff/math/maxs.xls
                          (there k+1 is chain length)

                          Best regards,

                          Andrey



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