Re: [PrimeNumbers] Re: Polynomials
- 2013/7/26 djbroadhurst <d.broadhurst@...>
> **Sure! But it (I suppose) does not satisfy the implicit assumption (explicit
> This polynomial performs better:
in the contest) that the primes must be consecutive.
[Non-text portions of this message have been removed]
- --- In firstname.lastname@example.org,
"WarrenS" <warren.wds@...> wrote:
> > Let N(a,n1,n2) be the number of primes of the formBad 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
> 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) = 148817to 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 :-)