This probably doesn't help much, but it seems like $p^{2p-1}||\text{disc}(F_p)$ for odd prime $p$.

@MichaelAlbanese the sum of $a_i$ is zero, so adding those $1/n$ is adding zero, but makes the infinite sums converge. The usage of the variable $x$ is traditional in partial fractions decomposition, indicating the equality is true for all, say, reals, and not just integers.

Ah you're right, I had an error in computation.

Ah, I see Maynard-Rudnick proved the upper bound part of the conjecture, which I think is enough for my 1/12 statement.

That there is a $\varepsilon>0$ for which we can disprove the statement for all smaller numbers, I think, follows from Cilleruelo's conjecture for the polynomial in question, and specifically we can take $\varepsilon=1/12$.

You will probably enjoy Gelman's and Nolan's foray into (2-d) golf. They wrote up the first part here stat.columbia.edu/~gelman/research/published/golf.pdf and added data here statmodeling.stat.columbia.edu/2019/03/21/… with further links there.

I think it should possible to find a specific curve over rationals, such that you can show that (up to log factors) a positive proportion of quadratic twists with discriminant supported on the first $n$ primes have root number -1, giving a much higher lower bound than logarithmic

@StanleyYaoXiao Don't we expect the average rank to increase with number field degree?

Ah! Yes! You're correct. Thanks!

Thank you for your question. I mean for all $i$, I have edited the question to reflect this