- 2013/7/26 Alexander <werner.sand@...>

> **

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

>

>

>

>

> I don't have your math program.

>

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

--

>

>

> --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...>

> 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] - --- In primenumbers@yahoogroups.com,

"WarrenS" <warren.wds@...> wrote:

> > Let N(a,n1,n2) be the number of primes of the form

Bad models.

> > 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,

No. Rather it is that n1, the begining of the sampling

> needs to be substantially larger than a, in order for naive

> statistical models to become reasonably valid?

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

to allow the HL heuristic to show through.

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

It's a simple as that. No shock-horror for statisticians;

Just a trivial observation by a log-lover :-)

David