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## Re: [PrimeNumbers] Re: Polynomials

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• 2013/7/26 Alexander ... That s Pari-GP, Alexander. You can get it here: http://pari.math.u-bordeaux.fr/ (It seems not to be working
Message 1 of 33 , Jul 27, 2013
2013/7/26 Alexander <werner.sand@...>

> **
>
>
>
>
> I don't have your math program.
>

That's Pari-GP, Alexander. You can get it here:

http://pari.math.u-bordeaux.fr/

(It seems not to be working right now).

Can I get your P(n) explicitly?
>
Do you really want us to paste here a polynomial of degree 99 with big
rational coefficients? To see what I'm saying, its independent term is
-11271426216601243783015961864589 !!

(But if you paste David's program on Pari and correct the missing }, you
will get a glimpse of it, if so you wish).

Jose

> Werner
>
>
> wrote:
> >
> >
> >
> > --- 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 :-)
> >
> > > 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, 2013
"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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