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RE: [PrimeNumbers] Consecutive primes

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  • Jon Perry
    This thread reminds me of an ancient conjecture by the great (and late) Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma(i=0 to oo;a_i.x^i)
    Message 1 of 16 , Apr 19, 2003
      This thread reminds me of an ancient conjecture by the great (and late)
      Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma(i=0 to
      oo;a_i.x^i) with integer coefficients a_i, a_i !=0 for at least one i!=0,
      always contains at least 1 composite.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths/
      http://www.users.globalnet.co.uk/~perry/DIVMenu/
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
    • j_m_berg
      Usual disclaimer: HS-dropout, no college, self-taught I ve no idea whether this little equation is already known, or whether it would even be of any interest.
      Message 2 of 16 , Apr 19, 2003
        Usual disclaimer: HS-dropout, no college, self-taught

        I've no idea whether this little equation is already known, or
        whether it would even be of any interest. But I was tinkering with
        some numbers and discovered something that is rather interesting to
        me.

        (3+6*N)^2+2

        This simple equation seems to yield a rather large number of primes.
        Enough so to where I've seen some rather large runs of consecutive
        primes (as many as 8 so far).
      • physiguy
        In reponse to Mr. Berg post, ... Well, one reason might be that: (3+6*N)^2+2 == 2 ~== 0 (mod 3) and, (3+6*N)^2+2 == 9*(1+2*N)^2+2 == 9*(1+2*N)^((5-1)/2)+ 2 ==
        Message 3 of 16 , Apr 19, 2003
          In reponse to Mr. Berg post,

          > I've no idea whether this little equation is already known, or
          > whether it would even be of any interest. But I was tinkering with
          > some numbers and discovered something that is rather interesting
          > to me.

          > (3+6*N)^2+2

          > This simple equation seems to yield a rather large number of
          > primes. Enough so to where I've seen some rather large runs of
          > consecutive primes (as many as 8 so far).


          Well, one reason might be that:

          (3+6*N)^2+2 == 2 ~== 0 (mod 3)

          and,

          (3+6*N)^2+2 == 9*(1+2*N)^2+2 == 9*(1+2*N)^((5-1)/2)+ 2
          == -1*(+-1) + 2 == +-1 + 2 ~== 0 (mod 5)

          so it's not divisble by 3 or 5. That and a little luck would
          probably help. I could analyze this more but I'll leave that up to
          you. But if you really want to help yourself, go back to school.

          Ehren
        • j_m_berg
          ... Thank you for your analysis and suggestion. But at age of 50, I think I ll continue dabbling and not spend the next half dozen years book-learning - as
          Message 4 of 16 , Apr 19, 2003
            --- In primenumbers@yahoogroups.com, "physiguy" <physiguy@h...> wrote:

            > you. But if you really want to help yourself, go back to school.

            Thank you for your analysis and suggestion.

            But at age of 50, I think I'll continue dabbling and not spend the
            next half dozen years book-learning - as long as that's okay with
            you, of course.
          • Jud McCranie
            ... It also isn t divisible by 13, 23, 29, 31, 37, 47.
            Message 5 of 16 , Apr 19, 2003
              At 09:43 PM 4/19/2003, physiguy wrote:

              >Well, one reason might be that:


              >so it's not divisble by 3 or 5.

              It also isn't divisible by 13, 23, 29, 31, 37, 47.
            • j_m_berg
              ... I could see originally from my own observation that it produced N s that were not divisible by small divisors, but I missed this point. Thank you for
              Message 6 of 16 , Apr 19, 2003
                --- In primenumbers@yahoogroups.com, Jud McCranie <judmccr@b...>
                wrote:
                > At 09:43 PM 4/19/2003, physiguy wrote:
                >
                > >Well, one reason might be that:
                >
                >
                > >so it's not divisble by 3 or 5.
                >
                > It also isn't divisible by 13, 23, 29, 31, 37, 47.

                I could see originally from my own observation that it produced N's
                that were not divisible by small divisors, but I missed this point.
                Thank you for bringing it out.
              • Jose Ramón Brox
                There are, implementing your formula in Pari-GP: 8 primes up to n=10 47 primes up to n=100 283 primes up to n=1000 2159 primes up to n=10^4 17476 primes up to
                Message 7 of 16 , Apr 20, 2003
                  There are, implementing your formula in Pari-GP:

                  8 primes up to n=10
                  47 primes up to n=100
                  283 primes up to n=1000
                  2159 primes up to n=10^4
                  17476 primes up to n=10^5
                  147150 primes up to n=10^6

                  A curious: if n is negative, say n=-k, then (3+6*n)^2 +2 = [3+6*(k-1)]^2 +2 (cuadratic <--> symmetric)

                  You can rewrite your formula like this:
                  (3+6n)^2 + 2 = 36n(n-1)+11

                  Taking k*n(n-1)+11, with the coefficients k=3,13,31 we have
                  308,405,370 primes up to 1000.

                  Amazingly your formula gets more primes than near all the other formulas with k primes (up to k=80).

                  I was thinking that all the k*n(n-1)+11 must have the same ratio primes/(count_of_n) as n becomes bigger, but maybe this isn't true. Anyone knows? For example, for n=10^5, for k=41, there are 10416 primes, which are over 7000 primes less than with k=36 (the k suggested by Berg). For k=31, we have 21719 primes for n=10^5.

                  Let's take again the formula: k*n(n-1)+11

                  With the squares:
                  k=1, n=1000 , primes: 289, n=10^5, primes: 15662
                  k=4, n=1000 , primes: 177, n=10^5, primes: 9884
                  k=9, n=1000 , primes: 206, n=10^5, primes: 12075
                  k=16, n=1000 , primes: 276, n=10^5, primes: 16698
                  k=25, n=1000 , primes: 213, n=10^5, primes: 13031
                  k=49, n=1000 , primes: 231, n=10^5, primes: 14381
                  k=64, n=1000 , primes: 234, n=10^5, primes: 14111
                  k=81, n=1000 , primes: 237, n=10^5, primes: 14760

                  PARI code
                  { k=? , j=? }
                  { psum=0;for(n=0,10^j,if(isprime(k*n*(n-1)+11),psum=psum+1));psum }


                  Let's search now for the five optimal k for n up to 10^3, 10^4 and for k up to 1000

                  I define the yield of the formula as number of primes / maximun n

                  n up to 10^3

                  k= 276, primes: 345. Yield: 0,345
                  k= 91, primes: 354. Yield: 0,354
                  k= 1000, primes: 361. Yield: 0,361
                  k= 31, primes: 369. Yield: 0,369
                  k= 13, primes: 404. Yield: 0,404

                  The average number of primes is 153.097 (counting multiples of 11 that trivially have only 1 prime, n=1)

                  The average yield is 0,153

                  Curios: 144*n*(n-1)+11 is always composite if n>1 (surely this holds for every M^2*n*(n-1)+(M-1) )

                  n up to 10^4

                  k= 133, primes: 2665. Yield: 0, 267
                  k=31, primes: 2693. Yield: 0, 269
                  k= 91, primes: 2722. Yield: 0, 272
                  k= 1000, primes: 2774. Yield: 0,277
                  k= 13, primes: 3033. Yield: 0,303

                  The average number of primes is 1198.

                  The average yield is 0,120

                  We observe that the numbers k are moreless the same. The other combinations are:
                  n=10^3
                  k=133, primes: 328. Yield: 0,328

                  n=10^4
                  k= 276, primes: 2601. Yield: 0,260

                  If we pick the first 10^4 numbers "as are", then by the PNT we will have about 10^4/log(10^4) = 1085 prime numbers. Using any of the k*n*(n-1)+11 formulas we have 10% more primes in average, and twice the amount if we make a good choice (like k=13) - yes, the sizes of the numbers are very different in both cases, i'm only supossing that we want to find N primes in at most P trys).


                  This problem remembered me the NumberSpiral of our friend Robert Sacks (http://www.numberspiral.com/index.html), because this formulas look like his P curves.

                  Now an (important?) question: if we take A*n(n-1)+B, what is the pair of numbers (A,B) that maximizes the amount of primes we get up to a maximum n? What about other non-linear formulas, like, for example, A*n(n-1)+B*n+C or A*n(n-1)*(n-7)+B ?

                  Is all this studied yet?

                  Jose Brox
                  http://espanol.groups.yahoo.com/group/Telecomunicacion



                  ----- Original Message -----
                  From: j_m_berg
                  To: primenumbers@yahoogroups.com
                  Sent: Sunday, April 20, 2003 2:27 AM
                  Subject: [PrimeNumbers] Consecutive primes


                  Usual disclaimer: HS-dropout, no college, self-taught

                  I've no idea whether this little equation is already known, or
                  whether it would even be of any interest. But I was tinkering with
                  some numbers and discovered something that is rather interesting to
                  me.

                  (3+6*N)^2+2

                  This simple equation seems to yield a rather large number of primes.
                  Enough so to where I've seen some rather large runs of consecutive
                  primes (as many as 8 so far).



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                  [Non-text portions of this message have been removed]
                • Jack Brennen
                  ... Sort of. See A17 in Richard Guy s Unsolved Problems in Number Theory. The asymptotic density of primes given by a quadratic with integer coefficients is
                  Message 8 of 16 , Apr 20, 2003
                    > Now an (important?) question: if we take A*n(n-1)+B, what is the pair
                    > of numbers (A,B) that maximizes the amount of primes we get up to a
                    > maximum n? What about other non-linear formulas, like, for example,
                    > A*n(n-1)+B*n+C or A*n(n-1)*(n-7)+B ?
                    >
                    > Is all this studied yet?

                    Sort of. See A17 in Richard Guy's Unsolved Problems in Number Theory.

                    The asymptotic density of primes given by a quadratic with integer
                    coefficients is believed to be c*sqrt(n)/log(n).

                    Of course, c can be as low as 0. An easy example is x^2+x+4, which
                    is never prime.

                    The value of c can be made high as well. Guy gives the "best"
                    example with the function:

                    x^2+x+132874279528931

                    for which he gives c = 5.0870883.

                    This polynomial is never divisible by any prime smaller than 181,
                    and asymptotically should yield many primes.

                    For x in the range [0 ... 10000], this polynomial yields 3141 primes,
                    which compares quite favorably with all of your examples.

                    Euler's well known function x^2+x+41 has corresponding c = 3.3197732,
                    and in the range [0 ... 10000] yields 4149 primes, quite a good count.
                    But the polynomial above should overtake it eventually.

                    Another function is x^2+x+27941, with c = 3.6319998. This one gives
                    an even more impressive 4466 primes in the range [0 ... 10000]. But
                    it too should eventually be surpassed by the one above.
                  • Adam
                    To address the mathematics, as long as p is a prime for which -2 is a quadratic residue and relatively prime to 6, p will divide (3+6*n) ^2+2 twice in a string
                    Message 9 of 16 , Apr 20, 2003
                      To address the mathematics, as long as p is a prime for which -2 is a
                      quadratic residue and relatively prime to 6, p will divide (3+6*n)
                      ^2+2 twice in a string of p consecutive values for n. Once you solve
                      x^2+2==0 mod p, you can then solve 3+6*n==x mod p. For this n value,
                      and every p thereafter (n+p, n+2p, n+3p...) the value will always be
                      divisible by p. Such a prime is p=11, for which 3^2=9==-2 mod 11.
                      Of course, the other root is (-3)^2==-2 mod 11. So, if n is 0 or 10
                      mod 11, (3+6*n)^2+2 is divisible by 11 (and if n is large enough,
                      this guarantees it is not prime). The longest possible string of
                      primes is therefore n==1,2,3,4,5,6,7,8,9 mod 11, and, with finding a
                      string of 8, you are almost to the predicted maximum.

                      Other such primes are 17, 19, 41, 43, 59, 67, 73, 83, 89, 97.
                      Perhaps the Chinese remainder theorem could be employed to limit the
                      length of possible prime sequences further, using this info.

                      To address the other topics, I support your curiosities, and urge you
                      to continue having fun with math topics. I would suggest that your
                      individual "educational status" is irrelevant, and you don't need to
                      post that at all. It is hard to ignore insensitive people, but it is
                      also a lot more wonderful world to live in when you remove their
                      influence from your life (as much as is possible).

                      Adam


                      --- In primenumbers@yahoogroups.com, "j_m_berg" <jberg@e...> wrote:
                      > Usual disclaimer: HS-dropout, no college, self-taught
                      >
                      > I've no idea whether this little equation is already known, or
                      > whether it would even be of any interest. But I was tinkering with
                      > some numbers and discovered something that is rather interesting to
                      > me.
                      >
                      > (3+6*N)^2+2
                      >
                      > This simple equation seems to yield a rather large number of
                      primes.
                      > Enough so to where I've seen some rather large runs of consecutive
                      > primes (as many as 8 so far).
                    • Jens Kruse Andersen
                      ... I only count probable primes in the following and may be one off because some results are old and started at x=1 - too lazy to check x=0 now.
                      Message 10 of 16 , Apr 20, 2003
                        Jack Brennen wrote:
                        > Sort of. See A17 in Richard Guy's Unsolved Problems in Number Theory.
                        >
                        > The asymptotic density of primes given by a quadratic with integer
                        > coefficients is believed to be c*sqrt(n)/log(n).
                        >
                        > Of course, c can be as low as 0. An easy example is x^2+x+4, which
                        > is never prime.
                        >
                        > The value of c can be made high as well. Guy gives the "best"
                        > example with the function:
                        >
                        > x^2+x+132874279528931
                        >
                        > for which he gives c = 5.0870883.
                        >
                        > This polynomial is never divisible by any prime smaller than 181,
                        > and asymptotically should yield many primes.
                        >
                        > For x in the range [0 ... 10000], this polynomial yields 3141 primes,
                        > which compares quite favorably with all of your examples.
                        >
                        > Euler's well known function x^2+x+41 has corresponding c = 3.3197732,
                        > and in the range [0 ... 10000] yields 4149 primes, quite a good count.
                        > But the polynomial above should overtake it eventually.
                        >
                        > Another function is x^2+x+27941, with c = 3.6319998. This one gives
                        > an even more impressive 4466 primes in the range [0 ... 10000]. But
                        > it too should eventually be surpassed by the one above.

                        I only count probable primes in the following and may be one off because some
                        results are old and started at x=1 - too lazy to check x=0 now.
                        x^2+16161x+6067 gives 4761 primes to x=10000.

                        Only going to 10000 gives small coefficients a big advantage, because the
                        candidates simply become smaller:
                        x^2+x+27941 gives 4466 primes to 10000 but only 35150 to 100000.
                        x^2+x+854032997 gives 4366 primes to 10000 and 41320 to 100000.

                        Try to beat my record in CYF NO. 13 at www.shyamsundergupta.com/canyoufind.htm
                        This allows negative and repeated primes and I got 44484 primes to 100000 in
                        x^2-100001x+2498695637
                        The puzzle asks for 50000 but that is beyond me.
                        According to my computations, 92.4% of the candidates will not have a prime
                        factor below 250 for x^2+x-7731189253. It only gave 41503 primes to 100000.

                        There are also results and discussions in this german paper:
                        www.birkhauser.ch/journals/1700/papers/9054002/90540064.pdf
                        Even if you don't know german, you may get something out of it.

                        --
                        Jens Kruse Andersen
                      • physiguy
                        ... Adam, That is an absolutely horrible thing to say, remove their influence from your life. Whether you realize it or not, everybody influences everybody
                        Message 11 of 16 , Apr 20, 2003
                          > It is hard to ignore insensitive people, but it is also a lot more
                          > wonderful world to live in when you remove their influence from
                          > your life (as much as is possible).

                          Adam,

                          That is an absolutely horrible thing to say, "remove their influence
                          from your life." Whether you realize it or not, everybody influences
                          everybody else in this world and that's what life is all about. And
                          you don't have any idea how sensitive I am. I am a math tutor, and I
                          haven't had a student yet that hasn't gotten an A. I was also a
                          cancer researcher at NCI and developed a computer program that
                          discovers new cancer treatment drugs. The one and only program that
                          is able to do this. So, if you get cancer you'll be glad you didn't
                          remove my influence from your life. I care about people
                          tremendously, and I've made it my personal and professional goal in
                          life to help as many people as I can. When I see somebody who looks
                          like they need help I'll do whatever I can to help them. Even if
                          that means rebuke. You know, rebuke is not a bad thing. What would
                          happen if you never scolded a child. I'll tell you what would
                          happen, they would never learn, they would never accomplish
                          anything, they would never have a happy and productive life, and the
                          list goes on. I certainly respect Mr. Berk and only want the best
                          for him. And I do not mean that in any way other than the way I said
                          it. When he refered to "book-learning" as if it was something he was
                          terrified of, I felt a need to help, whether or not he wanted my
                          advice. I saw a way I could help him and I and tried to do so. That
                          is all there was to it. There was no insensitivity ever directed
                          towards him or you. It would only be insensitive if I ignored his
                          situation. I hope this has cleared up my intentions. So, encourage
                          Mr. Berk to indepently study, but I also want to clarify that "book-
                          learning" is not a bad thing, no matter how old you are.

                          Ehren
                        • Jud McCranie
                          At 07:24 PM 4/20/2003, physiguy wrote: like they need help I ll do whatever I can to help them. Even if ... We have to remember, though, that there are many
                          Message 12 of 16 , Apr 20, 2003
                            At 07:24 PM 4/20/2003, physiguy wrote:

                            like they need help I'll do whatever I can to help them. Even if
                            >that means rebuke. You know, rebuke is not a bad thing.

                            We have to remember, though, that there are many types of people on this
                            list. Some are students, some have an interest in prime numbers but aren't
                            mathematically sophisticated, etc. For instance, a bright 12 year old
                            might make some observation about primes, and we wouldn't want to rebuke
                            them, would we?
                          • j_m_berg
                            ... I ve spent three decades as a systems design engineer in Silicon Valley and chances are I ve helped design much of the systems you ve programmed. ... Thank
                            Message 13 of 16 , Apr 20, 2003
                              > haven't had a student yet that hasn't gotten an A. I was also a
                              > cancer researcher at NCI and developed a computer program that
                              > discovers new cancer treatment drugs. The one and only program that
                              > is able to do this.

                              I've spent three decades as a systems design engineer in Silicon
                              Valley and chances are I've helped design much of the systems you've
                              programmed.

                              > like they need help I'll do whatever I can to help them. Even if
                              > that means rebuke.

                              Thank you for rebuking me. However who asked you to do so? I was
                              presenting something of interest to myself (and others, judging by
                              the thread). As I said before, it was for you to decide whether to
                              investigate it or ignore it.

                              > ... What would
                              > happen if you never scolded a child. I'll tell you what would
                              > happen, they would never learn, they would never accomplish
                              > anything, they would never have a happy and productive life, and
                              the
                              > list goes on.

                              At 50, I doubt I am a child. And I seem to have a happy and
                              productive life after three decades as a self-taught systems
                              engineer. However it now becomes obvious why you feel that you're
                              qualified to rebuke me - given that you view me as a child.

                              > it. When he refered to "book-learning" as if it was something he
                              was
                              > terrified of, I felt a need to help, whether or not he wanted my
                              > advice. I saw a way I could help him and I and tried to do so.

                              I've spent my life "learning" since I'm a self-taught systems
                              engineer. The fact that I choose not to spend the next half decade
                              getting a degree in math, is simply that I see no need to do so for a
                              hobby! Thank you for your "help", but your entire theory was
                              incorrect given that I obviously have zero fear of "learning". This
                              includes the dozens of courses I've taken on subjects that I have
                              choosen to pursue.

                              > is all there was to it. There was no insensitivity ever directed
                              > towards him or you. It would only be insensitive if I ignored his
                              > situation.

                              Personally I would have preferred you to ignore my "situation" if it
                              means you weren't so damned condescending! People don't like having
                              someone tell them that their hobby is equivilent to cloud watching!
                              While I obviously won't create any breakthroughs, it is my right to
                              enjoy my life as I wish!
                            • Adam
                              After being on this list for several months, I begin to understand others frustration with your post, Jon. Polynomials are not composite, rather, they are
                              Message 14 of 16 , Apr 22, 2003
                                After being on this list for several months, I begin to understand
                                others' frustration with your post, Jon. Polynomials are not
                                composite, rather, they are reducible or irreducible, and you really
                                need to also mention the ring that is attached to that status,
                                reducible over... or irreducible over.... Since you connected your
                                post to the thread, I would assume that you meant "a polynomial that
                                has a function value which is composite."

                                As to it being a conjecture, one of my previous post sketched a proof
                                that for infinity many primes p there are infinitely many values x
                                for which f(x) is divisible by p^2. So, to answer the conjecture,
                                yes, function values are composite, not just once but, infinitely
                                often.

                                See: Message 11921 of 12227

                                Adam

                                --- In primenumbers@yahoogroups.com, "Jon Perry" <perry@g...> wrote:
                                > This thread reminds me of an ancient conjecture by the great (and
                                late)
                                > Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma
                                (i=0 to
                                > oo;a_i.x^i) with integer coefficients a_i, a_i !=0 for at least one
                                i!=0,
                                > always contains at least 1 composite.
                                >
                                > Jon Perry
                                > perry@g...
                                > http://www.users.globalnet.co.uk/~perry/maths/
                                > http://www.users.globalnet.co.uk/~perry/DIVMenu/
                                > BrainBench MVP for HTML and JavaScript
                                > http://www.brainbench.com
                              • Adam
                                A slight modification in this polynomial yields a slightly higher count of primes. In searching for primes of the form ax^2+bx+c, it is good to put the axis
                                Message 15 of 16 , May 4, 2003
                                  A slight modification in this polynomial yields a slightly higher
                                  count of primes. In searching for primes of the form ax^2+bx+c, it
                                  is good to put the axis of symmetery at -b/2a=50000.5 so that when x
                                  yields a prime value then 100001-x also yields a prime value (what we
                                  Americans call a "two-fer," for two for one sales). The general
                                  principle is sound but one should realize that, once a candidate
                                  formula is found, the value at x=1 might be composite while the value
                                  at x=100001 might be prime, so that shifting x by 1 would up the
                                  prime count. In this fashion I identified a shift of 5357 giving 16
                                  more primes. The polynomial x^2-89287*x+1991687729 gives 44500 Prp
                                  on the range x=1..100000.

                                  Adam Stinchcombe

                                  snip>>>>

                                  Try to beat my record in CYF NO. 13 at
                                  www.shyamsundergupta.com/canyoufind.htm
                                  > This allows negative and repeated primes and I got 44484 primes to
                                  100000 in
                                  > x^2-100001x+2498695637
                                  > The puzzle asks for 50000 but that is beyond me.
                                  > According to my computations, 92.4% of the candidates will not have
                                  a prime
                                  > factor below 250 for x^2+x-7731189253. It only gave 41503 primes to
                                  100000.
                                  <<<snip
                                • Jens Kruse Andersen
                                  ... Nice work. I also considered trying to shift when I found the symmetric quadratic. I knew the requested 50000 primes was far out of reach and was too lazy
                                  Message 16 of 16 , May 5, 2003
                                    Adam wrote:
                                    > A slight modification in this polynomial yields a slightly higher
                                    > count of primes. In searching for primes of the form ax^2+bx+c, it
                                    > is good to put the axis of symmetery at -b/2a= so that when x
                                    > yields a prime value then 100001-x also yields a prime value (what we
                                    > Americans call a "two-fer," for two for one sales). The general
                                    > principle is sound but one should realize that, once a candidate
                                    > formula is found, the value at x=1 might be composite while the value
                                    > at x=100001 might be prime, so that shifting x by 1 would up the
                                    > prime count. In this fashion I identified a shift of 5357 giving 16
                                    > more primes. The polynomial x^2-89287*x+1991687729 gives 44500 Prp
                                    > on the range x=1..100000.
                                    >
                                    > Adam Stinchcombe

                                    Nice work. I also considered trying to shift when I found the symmetric
                                    quadratic. I knew the requested 50000 primes was far out of reach and was too
                                    lazy to try the shift. I did no heuristics but did not expect as many as 16
                                    extra primes.
                                    Brian Trial's best attempt at www.shyamsundergupta.com/canyoufind.htm is a
                                    shift of my second best, but he found it independently. It only has one extra
                                    prime and only shifts the symmetri axis from 50000.5 to 49999.5. Maybe he did
                                    not shift but started with a slightly asymmetric quadratic on the interval
                                    1..100000.

                                    --
                                    Jens Kruse Andersen
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