## Re: Polynomials

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• djbroadhurst wrte: ... Small supplement your results Pk - n = 0..10^k {rank/P4,{a, P2, P4, P5, P6}} {1,{247757,71,5028,39759,324001}}
Message 1 of 33 , Jul 26, 2013
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> Here are the top 20, ranked by
> P4 = sum(n=0,10^4,isprime(n^2+n+a))
> in the last column. I also give
> P2 = sum(n=0,10^2,isprime(n^2+n+a))
> which is not always a reliable guide
> to subsequent performance.
>
> rank a P2 P4
> [1, [ 247757, 71, 5028]]
> [2, [ 595937, 61, 4978]]
> [3, [ 1544987, 59, 4809]]
> [4, [ 2640161, 61, 4799]]
> [5, [ 2367767, 55, 4767]]
> [6, [ 239621, 62, 4724]]
> [7, [ 701147, 62, 4713]]
> [8, [ 115721, 69, 4691]]
> [9, [ 3106001, 46, 4684]]
> [10, [ 8930807, 55, 4676]]
> [11, [ 771581, 59, 4670]]
> [12, [ 333491, 60, 4669]]
> [13, [ 765197, 57, 4666]]
> [14, [ 1974881, 52, 4644]]
> [15, [ 361637, 63, 4634]]
> [16, [15102077, 46, 4631]]
> [17, [ 383681, 63, 4613]]
> [18, [ 722231, 57, 4610]]
> [19, [ 136517, 66, 4609]]
> [20, [ 601037, 57, 4605]]
>
> I believe that this list is complete for
> a <= 10^8 and P4 >= 4605.
>
Pk -> n = 0..10^k

{rank/P4,{a, P2, P4, P5, P6}}
{1,{247757,71,5028,39759,324001}}
{2,{595937,61,4978,39293,322141}}
{3,{1544987,59,4809,38361,314274}}
{4,{2640161,61,4799,38501,315542}}
{5,{2367767,55,4767,38058,312437}}
{6,{239621,62,4724,36976,303476}}
{7,{701147,62,4713,37376,305818}}
{8,{115721,69,4691,37059,302999}}
{9,{3106001,46,4684,37831,310304}}
{10,{8930807,55,4676,38343,315433}}
{11,{771581,59,4670,36404,299303}}
{12,{333491,60,4669,36592,300002}}
{13,{765197,57,4666,36893,303389}}
{14,{1974881,52,4644,37261,305096}}
{15,{361637,63,4634,36644,299939}}
{16,{15102077,46,4631,38561,318251}}
{17,{383681,63,4613,35863,294775}}
{18,{722231,57,4610,36610,299667}}
{19,{136517,66,4609,36230,297046}}
{20,{601037,57,4605,36308,297251}}

--
marian otremba
• ... 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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