## Re: Polynomials

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• I sent my reply wrong, I m afraid, so once more: I collect (each case n=1,2,3..100, I cannot say per cent): n²-n+41 (Euler) generates 86 primes n²-n+27941
Message 1 of 33 , Jul 26 5:45 AM
I sent my reply wrong, I'm afraid, so once more:

I collect (each case n=1,2,3..100, I cannot say per cent):

n²-n+41 (Euler) generates 86 primes
n²-n+27941 generates 77 primes
¼* (n^5-133n^4+6729n^3 -158379n^2+1720294n-6823316) generates only 64 primes

I checked n²-n+a up to a=10^6, no better result. Can anybody compute further? Can a better result actually be expected?

Werner

--- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@...> wrote:
>
> Now that I think about it better, I see that I was fooled by your proposal!
> :D
>
> What we want is not a polynomial whose _roots_ are primes, what we want is
> a polynomial which generates as much primes as possible with its _image_.
> I.e., what 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.
>
> Regards,
> Jose Brox
>
>
> 2013/7/26 Jose Ramón Brox <ambroxius@...>
>
> >
> >
> >
> >
> >> **
> >> This polynomial performs better:
> >>
> >> polinterpolate(vector(100,k,prime(k)))
> >>
> >>
> > Sure! But it (I suppose) does not satisfy the implicit assumption
> > (explicit in the contest) that the primes must be consecutive.
> >
> > Regards,
> > Jose
> >
> >
>
>
> --
> La verdad (blog de raciocinio político e información
> 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 8:25 AM
"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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