- 2013/7/27 Jose Ram�n Brox <ambroxius@...>

>

What I really wanted to say is: if you paste David's _polynomial_ and add

> (But if you paste David's program on Pari and correct the missing }, you

> will get a glimpse of it, if so you wish).

>

>

the _necessary_ }, you will get it. You must also erase the ";", like this:

{Poly=polinterpolate(vector(100,k,prime(137*k-104)))}

> 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/>

>

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