Date: Oct 1, 2017 11:46 AM Author: David Bernier Subject: Prime mystery in Euler's polynomial P(k) := k^2 -k + 41

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?

Mystified,

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

-----------------------------------------------------------