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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: 3,892 Registered: 12/13/04
Prime mystery in Euler's polynomial P(k) := k^2 -k + 41
Posted: Oct 1, 2017 11:46 AM

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

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

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