
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 primegenerating polynomial:
up to a simple change of variable (unitshift).
almost, i.e.
gg(Q1) = 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 coprimeness 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


