@domotorp, it's not monotonic, because e.g. the ratio for $n=15$ is larger than $n=14$, but it is monotonic between $n=15$ and $n=23$ so that could potentially be the last exception.

@domotorp, Python code and values to n=23. I must say that it doesn't look to me to be converging to a constant.

It's on archive.org (https://archive.org/details/combinatorialide00john). Requires a free account to access it, and it's a library with restrictions on simultaneous borrowing, but I've used this online copy when chasing references in the past.

Fair point. I was processing an offline database which only has the sequence number and the entries. I can't remember offhand whether it's possible to download one which has the keywords in order to identify triangles and process them specially. If not, I could look at trying to break sequences into triangles and maybe deal with a few false positives. I suppose it might also be worth trying to do the same with potential tabl sequences given by anti-diagonals.

With respect to rarity, I'm still manually filtering OEIS sequences to ones which are at least tangentially related to combinatorics, but pre-manual-filtering there are about six times as many plausibly non-trivially unimodal as plausibly non-trivially upside-down unimodal sequences in OEIS. (My concept of non-trival here requires at least three descents and at least three ascents, and the "plausibly" disclaimer is because my code only looks at the terms of the sequence available in the database).

Intuitively, it seems that the rarity of odd elements can be explained by the fact that two thirds of Fibonacci numbers are odd. This gives a bias towards even numbers at the start, and that bias then self-reinforces.

Have you tried exploiting the known values $f(2N, N-1) = 2^{2N-1} - \frac{1}{2} \binom{2N}{N}$ and $f(2N + 1, N) = 2^{2N}$ and handling central ranges by subtracting from / adding to these offsets?

Also, I'm attempting the check of paragraph 1 and failing. E.g. in $\mathbb{F}_3$ we have $Q = \{2\}$ and $Q^c = \{0\}$ but $2\cdot 2^2 - 1 = 1 \not \in Q$ and $2 \cdot 0^2 - 1 = 2 \not\in Q^c$.

It might be clearer if the polynomial and the prime weren't both called $p$.

So the pentagonal hexecontahedron?

Not forests: elements in the poset formed by edge contractions. And yes, for small graphs one can calculate. E.g. of the simple cubic graphs on 8 vertices two are non-planar and the others have distributions [(2, 39), (3, 146), (4, 207), (5, 146), (6, 58), (7, 12), (8, 1)], [(2, 54), (3, 220), (4, 283), (5, 176), (6, 62), (7, 12), (8, 1)], [(2, 63), (3, 268), (4, 345), (5, 202), (6, 66), (7, 12), (8, 1)].

Surely it depends on the triangulation, so what would your variables for the closed form be?

@OscarLanzi, the certificate is for a different domain. Looks like they're migrating, because the page is also present on the domain covered by the certificate: cs.smu.ca/~dawson/images4.html#Swirl

I think this can be tackled as a Markov process with five states: one for rejected pairs and one for each combination of true and approximate carry digit, processing from least significant.