- 2013/7/26 djbroadhurst <d.broadhurst@...>

> **

Yeah, you are so right! Hehe

>

>

>

>

> --- 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 :-)

>

>

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] - --- 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