I tried to compute the multiplicative inverse of the probability that $ 2 x^2 +1 $ is prime.
(I'm aware that proving there are infinitely many such primes is not done yet, but let's ignore that for now).
I know it should be around $2 \log(x) - 1$ from the PNT.
Using the sieve $(1-1/2)(1-1/3)(1-1/5) ... (1-1/p)$ over the primes $p$ we get $O(\log(x))$ actually $2C\log(x)$ with $C$ mertens constant. This is slightly worse but $C$ is not so far from $1$.
However when I tried differently I got confused.
I tried to sieve by using quadratic residues.
Let $p_n*$ be a prime such that it has a solution $x$ to $2 x^2 + 1 = 0 \pmod p$.
I believe that probability is about $1/2$.
Then I sieve : $(1-1/2)(1-1/p_1*)(1-1/p_2*)...(1-1/p_n*)$
And consider the worst case and average scenarios.
This does not agree well with $2 \log(x)$ even after dividing by $C$!??
What did I do wrong?