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• ... Indeed you were :-) ... for /consecutive/ values of the integer n in P(n), as per Euler. ... Again, better polynomials are trivial to construct. For
Message 1 of 33 , Jul 26
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Jose RamÃ³n 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
• ... 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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