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Re: Two large consecutive smooth numbers

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  • djbroadhurst
    ... Oh well, I guess that, having been set up by Phil, I ought not to pass. A quick coding of my favourite PTE identity seemed to leave a significant
    Message 1 of 17 , Mar 3 8:14 PM
      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > Warren knows about Prouhet-Tarry-Escott and
      > has set a puzzle that goes deeper than that.
      >
      > I pass.

      Oh well, I guess that, having been set up by Phil,
      I ought not to pass. A quick coding of my favourite
      PTE identity seemed to leave a significant computational
      load for Pari-GP. I am willing to let that code run
      for a few hours, without taking the effort to tune it.

      David
    • djbroadhurst
      ... but might, less coyly and more helpfully, have written {m=67440294559676054016000;y=1094090867210^2;
      Message 2 of 17 , Mar 3 9:27 PM
        --- In primenumbers@yahoogroups.com,
        "WarrenS" <warren.wds@...> wrote:

        > N=43623575184339996059537425773119366447006380455838\
        > 696504055889999185302903791148393125043181272726633463298672436846034128

        but might, less coyly and more helpfully, have written

        {m=67440294559676054016000;y=1094090867210^2;
        N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-1;}

        in the more explicit manner of
        http://physics.open.ac.uk/~dbroadhu/cpte.pdf

        David
      • 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 3 of 17 , Mar 4 6:37 AM
          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 4 of 17 , Mar 4 6:59 AM
            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 5 of 17 , Mar 4 7:45 AM
               
              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 6 of 17 , Mar 4 10:56 AM
                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 7 of 17 , Mar 4 10:58 AM
                  --- 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 8 of 17 , Mar 4 11:35 AM
                    --- 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 9 of 17 , Mar 4 1:40 PM
                      --- 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 10 of 17 , Mar 5 12:48 AM
                        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 11 of 17 , Mar 5 1:04 AM
                          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 12 of 17 , Mar 5 11:28 AM
                            > 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 13 of 17 , Mar 5 4:09 PM
                              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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