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25270Re: [PrimeNumbers] Re: Polynomials

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  • Jose Ramón Brox
    Jul 27, 2013
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      2013/7/26 djbroadhurst <d.broadhurst@...>

      > **
      >
      >
      >
      >
      > --- In primenumbers@yahoogroups.com,
      > Jose Ram�n Brox <ambroxius@...> wrote:
      >
      > > Now that I think about it better,
      > > I see that I was fooled by your proposal!
      >
      > Indeed you were :-)
      >
      >
      Yeah, you are so right! Hehe

      You can see that I don't know that much Pari-GP programming; I thought that
      "polinterpolate" of the vector V was interpolation at (V, [0 ... 0]), but
      it is actually at ([1 2 ... length(V)], V)!

      In our contest (from Al Zimmermann's Programming Contests) we had a prize
      for every degree from 1 to 10, so the problem had a constraint on the
      degree (which is, in my opinion, the sensible way to tackle the problem;
      otherwise there are trivial solutions as you suggest).

      The rules, submitted solutions and winners of that contest can be found
      here:

      http://www.recmath.org/contest/PGP/

      You also have a routine that will tell you the scoring as it was in the
      contest if you were to send an specific polynomial.

      Could you make it better than the winners? :D

      Rgards,
      Jose




      >
      > > what we want is a polynomial which generates as much primes
      > > as possible
      >
      > for /consecutive/ values of the integer n in P(n),
      > as per Euler.
      >
      >
      > > we want is to maximize p such that the sequence p({1,2,3,...})
      > > gives the biggest possible "prime head".
      >
      > > The classical example of Euler, p(x)=x^2-x+41, e.g., gives p(1)=41,
      > > p(2)=43, p(3)=47, and so on.
      >
      > Again, better polynomials are trivial to construct. For example
      > here is a polynomial P(n) whose image gives 100 primes
      > for n = 1 ... 100, starting with P(1) = 137
      >
      > P(n)=subst(Poly,x,n);
      > {Poly=polinterpolate(vector(100,k,prime(137*k-104)));
      > for(n=1,100,image=P(n);if(!isprime(image),print(fail),
      > print("P("n") = "image)));}
      >
      > P(1)=137
      > P(2)=1013
      > P(3)=2027
      > P(4)=3119
      > .....
      > P(97)=142007
      > P(98)=143719
      > P(99)=145463
      > P(100)=147073
      >
      > Amusingly, the next prime is
      >
      > P(137) = 7697722242995617000095454995366617089022284749995846581728875603
      >
      > David
      >
      >
      >



      --
      La verdad (blog de raciocinio pol�tico e informaci�n
      social)<http://josebrox.blogspot.com/>


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