I read this in an older thread I cannot reply to:

Would even the SoB candidates converge to a similar efficiency as any other k in the end?

Yes, this is a conjecture. Today no statistical deviation was found to disprove it.

the weight of a multiplier k, after adjusting for algebraic factorizations, is empirically a good predictor of how frequently that k produces primes.

Yes, some k's produce fewer primes than others. (At least empirically. Basically nothing has been proven about how many primes any given k produces, except in some cases when the answer is 0.) But this discrepancy appears to be well-explained in all cases by how many candidates are sieved out (by small prime factors or algebraic factorizations). So the density of primes *among candidates that remain after sieving* doesn't seem to depend on k in the long run.

A simple, made-up example: Let's say you're testing two k's, "k1" and "k2", for primes up to n=1M. Let's say k1 has a low weight, like SoB k's, and k2 has a high weight. Then you may find that after sieving, you're left with 3000 candidates for k1 and 30000 candidates for k2, out of 1000000 each. When you run all the primality tests, it turns out that k1 produces 5 primes and k2 produces 50. So the higher-weight k, k2, produced more primes. But it didn't produce primes more "efficiently", because in order to find 10 times more primes, you needed to run 10 times more LLR tests. What Yves was saying in your first quote is that we don't expect any k to produce more or fewer primes per LLR test than any other k.

Ok, so the k's have different prime density but since the ones with lower density also produce less candidates, it evens out?