- 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@...>

>

> >

> >

> >

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

> >

> >> **

> >> 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]

> - --- 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