## Polynomials

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• http://groups.yahoo.com/group/primenumbers/files/Articles/polynomials.pdf (51k) There s an interesting polynomial sequence with easy-factorable coefficients,
Message 1 of 33 , Nov 2 2:18 PM
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There's an interesting polynomial sequence with easy-factorable coefficients, which I found two years ago when studying Lucas sequence for Mersenne prime test (4, 14, 194, ...).
These polynomials seem to have something general with Lucas polynomials of degree 2^n. Maybe it's the way to prove my two-year-old conjecture described in the paper?

Andrey

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• ... 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, 2013
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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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