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Topic: Prime mystery in Euler's polynomial P(k) := k^2 -k + 41
Replies: 3   Last Post: Oct 4, 2017 9:50 AM

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 David Bernier Posts: 3,884 Registered: 12/13/04
Re: Prime mystery in Euler's polynomial P(k) := k^2 -k + 41
Posted: Oct 1, 2017 4:42 PM

On 10/01/2017 11:46 AM, David Bernier wrote:
>
> gg(X):= X^2+X+41
>
> gg(.) is Euler's prime-generating polynomial:
>
> up to a simple change of variable (unit-shift).
>
>
> almost, i.e.
>
> gg(Q-1) = Q^2 - Q + 41 , [ Euler's polynomial in Q ].
> which is of the same form as Euler's k^2 - k + 41
> from Euler's Lucky numbers:
>
> < https://en.wikipedia.org/wiki/Lucky_numbers_of_Euler > .
>
> Modulo 41, two residue classes , k == 0 (mod 41)
> and  k == 1 (mod 41) yield a k^2 - k + 41 == 0 (mod 41).
> If k > 40, and either k == 0 or k == 1 (mod 41), then
> 41 divides k^2 -k +1, and this last number is not
> a prime.
>
> There remain 39 residue classes modulo 41 which aren't
> forbidden from producing primes, when k > 40.
>
> For "large" swaths of consecutive integers,
> I tested candidates, where a candidate,
> in terms of Euler's P(k) = k^2 - k+1,
> is a k>40 with k =/= 0 , k =/= 1 (modulo 41).
>
> These candidates are not divisible by 41.
>
> If K^2 - K +1 is prime, I give it a weight
> of log(K^2 - K +1).
>
> Then, I look at the sum of the weights of
> the primes of the form: k^2 - k+1,
> and the number of candidates, for k
> in a large range of consecutive integers
> [ a, b].
>
> I calculate the quotient:
>  (sum of weights of primes)/(number of candidates),
> for large intervals [a, b].
>
> This quotient approaches 6.98 over ranges [a, b]
> that include thousands to millions of candidates
> that are in fact probable primes (pseudoprimes).
>
> Example:
>
> For the range
> [ 3,000,000,001 ... 4,000,000,000]
>
> there are:
> 951,219,512 candidates X such that
> X^2+X+41 =/= 0 (mod 41)
>
> and there are:
>
> 151,101,437 pseudoprimes (probable primes),
>
> and the weight of the probable primes is
> 6,640,090,792.4
>
> and weight/candidates ~= 6.98 .
>
>
>
> I looked for patterns in prime factors of
>  x^2 + x + 41, when x^2 + x + 41 is composite,
> and found no pattern. [ equivalently, poly. k^2 - k + 41 ].
>
> So I'm puzzled as to why this 6.98 ~= 7 persists,
> even with x (or k) into a few billions.
>
> Could it all be explained by
> co-primeness to the primes from 2 to 37 inclusive?

I found a seminar handout by Edray Goins ( Purdue Math Dept)
on quadratic polynomials in Z[x] and a Conjecture of
Hardy and Littlewood known as "Conjecture F" having to
do with the distribution of prime numbers among the
values taken by f(x) = a*x^2 + b*x + c, for
a, b, c in Z.

He uses the notation pi_f (x), which I'm guessing
is

pi_f(x) := | { y: 1 <= y <= x, and f(y) prime } |

A link to his seminar handout in PDF format is the
following:

< http://www.math.purdue.edu/~egoins/seminar/12-12-07.pdf > .

Euler's quadratic appears in the form f(x) = x^2 + x + 41,
with disciminant Delta = -163.

A constant C(Delta) is defined by an infinite product involving
primes, the Legendre symbol, and -163 (or Delta).

Goins writes that H.C. Williams finds

C(-163) ~= 3.3197732

That's what I got evaluating a partial product with the PARI/gp
calculator, for the record:

pp=1.0;for(X=3,10^8,if(isprime(X),p=X;pp=pp*(1-kronecker(-163,p)/(p-1))));pp

= 3.3197732923520903506343596548256201696

If pi_f(x) is what I think it is, the Conjecture would say that

pi_f (x) ~= C(-163) li(x)

or pi_f(x) ~= 3.319 li(x) ,

for f(x) := x^2 + x + 1.

Then, according to my calculations, to get the
supposed constant ~= 6.98 in the computations below,

one takes:

(41/39)*2*C(-163) ~= 6.9800 .

I was looking for more references.

There's a web page with references, some encyclopedia:

"Hardy-Littlewood conjecture F",

< http://www.gutenberg.us/articles/hardy-littlewood_conjecture_f >

The imprint isn't great: bad-looking math symbols, elided references;
but it's from a physical book, I'm assuming.

David Bernier

>
> ---------------------------------------------------------------
>
> ? K=0
>  = 0
>
>
> for(D=3,10, summ = 0.0; count=0; np=0;
>     for(Z=K+1,K+10^D,bb=gg(Z);
>          if((bb%41)>0,count=count+1;
>             if(ispseudoprime(bb),np=np+1;summ=summ+log(bb))));
>
>     print(D,"   ",count,"  ",np,"  ",summ/count)  )
>
>
> 3   952  581  7.011
> 4   9514  4148  7.022
> 5   95122  31984  6.987
> 6   951220  261080  6.981
>
> ----------------------------------------------------------
>
> ? K
>  = 1000000000
>
>
> for(D=3,10, summ = 0.0;count=0;np=0;
>     for(Z=K+1,K+10^D,bb=gg(Z);
>          if((bb%41)>0,count=count+1;
>              if(ispseudoprime(bb),np=np+1;summ=summ+log(bb))));
>
>     print(D,"   ",count,"  ",np,"  ",summ/count)  )
>
>
> 3   950  159  6.936
> 4   9512  1625  7.08
> 5   95122  16083  7.007
> 6   951218  160439  6.990
>
> --------------------------------------------------------
>
> ? K = K + 10^9
> = 2000000000
>
>
>
>
> for(D=3,10, summ = 0.0;count=0;np=0;
>     for(Z=K+1,K+10^D,bb=gg(Z);
>          if((bb%41)>0,count=count+1;
>             if(ispseudoprime(bb),np=np+1;summ=summ+log(bb))));
>
>     print(D,"   ",count,"  ",np,"  ",summ/count)  )
>
>
> 3   950  166  7.48
> 4   9512  1477  6.65
> 5   95122  15499  6.979
> 6   951218  154943  6.977
> 7   9512194  1549537  6.978
>
> ------------------------------------------------------
>
> ? K = K + 10^9
>  = 3000000000
>
> for(D=3,10, summ = 0.0;count=0;np=0;
>     for(Z=K+1,K+10^D,bb=gg(Z);
>          if((bb%41)>0,count=count+1;
>             if(ispseudoprime(bb),np=np+1;summ=summ+log(bb))));
>     print(D,"   ",count,"  ",np,"  ",summ/count)  )
>
> 3   950  145  6.66
> 4   9512  1530  7.02
> 5   95122  15287  7.01
> 6   951220  152132  6.98
> 7   9512194  1,521,757  6.98
> 8   95121950  15,202,323  6.98
> 9   951,219,512  151,101,437  6.98
>
> -----------------------------------------------------------

Date Subject Author
10/1/17 David Bernier
10/1/17 David Bernier
10/2/17 David Bernier
10/4/17 David Bernier