Re: [PrimeNumbers] Polynomials
- The answer to both (1) and (2) is yes, if you believe
the first Hardy-Littlewood conjecture.
There is undoubtedly a value of (a) which will produce
100 different prime numbers for n=1..100. You're
basically describing an admissible 100-tuple. (It's
provably admissible because n^2-n cannot cover all
residues for any prime modulus.)
Naming such a value for (a) is currently out of reach;
we're not even ready to systematically search for
50-tuples with any chance of success, much less 100-tuples.
On 7/25/2013 9:37 AM, Alexander wrote:
> Let us have a look at the polynomial P=n²-n+a. a=41 results in the
> well-known Euler-formula, which produces 86 different prime numbers for
> n=1..100. The second-best a seems to be 27941 with 77 different primes.
> Two questions:
> 1) Is there a more efficient a (except 41)?
> 2) Is there another polynomial form which is more efficient? (no
> repeats, no negative results)
> [Non-text portions of this message have been removed]
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- --- 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 :-)