"If you want an algorithm that is guaranteed to solve the problem, assuming a prime pair exists, then you may need to provide sufficient computational resources to search every prime up to 𝑛/2" — this leaves me curious; assuming Goldbach's conjecture, is there anything known about the behavior of a 'minimal' Goldbach pair? e.g., can we prove that if it exists then for any $\epsilon$ there's always a pair $p+q$ with $p\lt o(n^\epsilon)$?

@FedorPetrov That sounds like a proof! Would you like to write it up or shall I?

How far out have you computed the $f_p()$? Since all the coefficients are positive and the leading and constant term are both $1$, your only possible rational root is $-1$, but I don't see any immediate proof from that.

(Side note: the name '$B$' is used in a couple of places where '$C$' should be, specifically in the paragraph(s) immediately after the statements of Lemmata 1 and 2. I would propose an edit but MO says that it's too small.)

Thank you! This is a really illuminating answer.

These make sense to me. Thank you!

I would think that negation could be defined as $f_{-r}(q) = \neg f_r(-q)$, and then subtraction by the formula above. And I'm not sure that my definition needs the fact that $V$ has no minimum; if there is a minimum element $q$ with $f_r(q)=\true$ then we just have $r=q$ (and this does mean there are two functions corresponding to every rational number, but that seems straightforwardly handleable).

I think Geoffrey means that the two planes of the triangles are very close to each other, and probably that the triangles' centers specifically are very close to each other.

If it helps any, two smooth $u()$ that satisfy your conditions and might be analyzable are $u(x)=1-\frac1{2(x^2+1)}$ and $u(x)=1-\frac12e^{-x^2}$.

@Simon see Andrej Bauer's answer for more specifics; the key notion is that you have a very small number of possible factors to check to see if $N$ is a prime power, that list of factors can be generated 'quickly' (polynomial in $\log N$), and the trial divisions can also be done quickly.

@IosifPinelis I think it's an excellent question and have certainly not downvoted/closevoted, but I suspect the folks who are are hoping for more context; it's clear that this polynomial comes from somewhere rather than appearing like Venus from the waves, and it would be helpful/interesting/etc. to know its origins.

I'm almost certainly missing something obvious, but isn't the Vornoi diagram of a collection of identical discs exactly the Vornoi diagram of their centers? The distance from any exterior point to a disc is the distance from that point to the center of the disc minus the disc's radius, so when trying to figure out the Vornoi cells all the points will have the same 'offset' applied to their distances and none of the comparisons will be changed.

I feel like the phrase 'the Chudnovsky formula for π is more attractive to mathematicians and computer scientists for computation due to its exotic and beautiful appearance' very much undersells how efficient the formula is for the problem. It's not the 'exotic and beautiful appearance' that's kept it in use for 35 years now, it's the effectiveness.

@anixx Yes and no. $\aleph_1$ is a cardinality; it is the cardinality of $\omega_1$ (the set of all countable ordinals). $\omega_1$ is an ordinal; it implies a well-ordering of the elements less than itself. But as a for-instance, assuming the Axiom of Choice the cardinality of the reals is at least $\aleph_1$ but we cannot extract an increasing sequence of order type $\omega_1$ from it.

This is an excellent answer; thank you. @Anixx Even assuming CH it makes little sense to me that you can get the ordinal $\omega_1$ from the cardinal $\aleph_1$, particularly since integration should use no potential well-ordering properties of the reals.

For clarification's sake, the definition of derivation you're using is the one from arxiv.org/abs/1503.00315 ?