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@prosfilaes, and from the other direction, if someone manages to prove that the I3 rank-into-rank axiom is necessary to prove the existence of a Laver table with period 32 (currently it's known to be sufficient and unless I've missed a recent proof there's no known proof with weaker assumptions) then that establishes an upper bound of BB(63) being knowable with weaker assumptions.
@MichaelHardy, slightly. $f(2n) = p_n$, $f(2n+1) = \lfloor \frac{p_n + p_{n+1}}{2}\rfloor$ needs slight tweaking to account for the sole prime gap of 1, but the idea should be clear.
Re "I do need to have some sense of what kinds of computations are feasible": there's an element here of knowing when to consult people with more experience in computation. I can think of one example where experience in combinatorial search reduces a calculation from a month to under a minute, and a few other examples which similarly show the value of better algorithms and more experience in optimisation.
No. The set of side length sequences is countable, but varying the angles between them may give an uncountable set for a single side length sequence. E.g. there's an uncountable set of unit rhombi.
You're looking at the output for $n = 6$. Change the third line (n = 6) for different values of $n$. If I'm reading Out[11] correctly in your output, you don't seem to be constraining f[0, 2] and f[2, 0] except to be equal to each other.
Ah, I've found the discrepancy. I misread the question as what @user675763 surely intended, which is to replace all of the $< N - 3$ with $\le N - 3$ so that every cell in the triangle has a constraint, rather than having an antidiagonal at $i + j = N - 3$ which is unconstrained.
@user11566470, what do you mean "there's no good implementation of integer-based Bin"? It's perfectly easy to implement. In C-like pseudocode, result = 1; for (int i = 0; i < k; i++) result = result * (n - i) / (i + 1);
We have $n^2$ elements to divide among $p$ residues, so there must be a multiset of at least $n^2 - p$ elements which can be partitioned into two equal multisets. If $n^2 - p \ge 2n$ then this means there must be a permutation with two equal rows, hence singular; so necessarily $n(n-2) < p$.
I think what this shows is either that your heuristic about $\exp$ and roots is wrong or that because we're working with formal power series there are convergence issues. From the definition of $b_i$ as $\sum_{i \ge -1} b_i z^i = \frac{\textrm{d}}{\textrm{d}z} \log f(z)$ we easily derive $f(z) = \exp \int \sum_{i \ge -1} b_i z^i \textrm{d}z = \exp \left(b_{-1}\log z + \int \sum_{i \ge 0} b_i z^i \textrm{d}z\right) = z^{b_{-1}} \exp \int \sum_{i \ge 0} b_i z^i \textrm{d}z$
The expression $\Pi(z) = z^2 \exp \sum_{i \ge 0} \frac{c_i z^{i+1}}{i+1}$ can also be derived directly from the definition of A307977 and the relationship $\Pi(z) = \frac{1}{1-z} \sum_{p \textrm{ prime}} z^p$.
