- djbroadhurst wrte:>
> Here are the top 20, ranked by

Small supplement your results

> 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 - --- 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