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).
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?