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• ## Re: [PrimeNumbers] Re: Polynomials

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• 2013/7/26 djbroadhurst ... Yeah, you are so right! Hehe You can see that I don t know that much Pari-GP programming; I thought that
Message 1 of 33 , Jul 27
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> **
>
>
>
>
> --- 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/>

[Non-text portions of this message have been removed]
• ... Bad models. ... No. Rather it is that n1, the begining of the sampling interval, needs to be substantially larger than sqrt(a), for the HL heuristic to win
Message 33 of 33 , Jul 28
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"WarrenS" <warren.wds@...> wrote:

> > Let N(a,n1,n2) be the number of primes of the form
> > n^2+n+a with n in [n1,n1+n2]. Then the data
> >
> > N(247757,0,10^6) = 324001
> > N(3399714628553118047,0,10^6) = 251841
> >
> > seem to favour the smaller value of a. Yet these data
> >
> > N(247757,10^12,10^6) = 148817
> > N(3399714628553118047,10^12,10^6) = 193947
> >
> > indicate that the larger value of a is better, in the long run.
>
> --these numbers seem to be in vast violation of naive statistical
> models.

> Is the reason, that the length n2 of the sampling interval,
> needs to be substantially larger than a, in order for naive
> statistical models to become reasonably valid?

No. Rather it is that n1, the begining of the sampling
interval, needs to be substantially larger than sqrt(a),
for the HL heuristic to win out. Clearly when
n1 < sqrt(3399714628553118047), Marion was comparing apples
and oranges, since log(n^2+n+a) was dominated by "a".

All I did was to level the playing field, here:

> N(247757,10^12,10^6) = 148817
> N(3399714628553118047,10^12,10^6) = 193947

to allow the HL heuristic to show through.

It's a simple as that. No shock-horror for statisticians;
Just a trivial observation by a log-lover :-)

David
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