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@Gro-Tsen, each of the entries satisfies a second-order D-finite recurrence with coefficients of degree $k$, so for some meanings of "general formula" it's not so hopeless. We have $$A_1 A_2 \cdots A_n = A_1 A_2 \cdots A_{n-1} \begin{pmatrix} (\frac{n}{n-1})^k & 0 \\ 0 & 1 \end{pmatrix} + A_1 A_2 \cdots A_{n-2} \begin{pmatrix} n^k & 0 \\ 0 & (n-1)^k \end{pmatrix}$$
If you consider a candidate assignment $\ell'(x)$ given by the majority function over the neighbours in $Y$, any deviation from that candidate assignment has strong implications. If $x_i$'s neighbours are assigned $0,1,1$ and we instead were to assign $0$ to $x_i$ then that forces (in general) four other values of $x_j$ to $1$. They may in turn force more values. It might be worth experimenting with medium-sized graphs to estimate how many values can be determined on average by searching for contradictions in these implication chains, which takes quadratic time.
Expanding as formal power series in $q$ and comparing constant terms gives $\Pi_1(a,b,c,d) = -\Pi_2(a,b,c,d)$. I don't think you're complicating things unnecessarily, so this makes me wonder whether there's a typo or a missing term.
Looks interesting. With $m=5$ it seems that the nim value of a pile of $n$ is eventually periodic with period 13. With $m=6$ the period appears to be 28.
"Minimal" is problematic. Firstly, the question of whether two TMs compute the same function is uncomputable. Secondly, you point out that the busy beaver function has a limit $N$ such that $BB(N)$ is independent of ZFC. This means that if you have a TM larger than $N$ which computes a function, you can't test all smaller TMs to see whether they compute the same function. FWIW I wouldn't be surprised if $N < 100$. I have a 64-state TM whose termination is currently only known, AFAIK, if we assume a rank-into-rank cardinal.
The circumcircle of three of the points lies in the surface of the circumsphere, so drop perpendiculars through two of the circumcentres and intersect them.
Are you using imaginary to mean non-real (in which case the answer does agree with the claim in the question) or real multiplied by $i$ (in which case it's stronger than the claim in the question)?
I have an example of this in a pre-print. The goal of the paper was to determine which abelian groups had a certain property. The proof ended up using a much weaker but more technical condition than "abelian group", and in particular worked for all quasigroups. My coauthor persuaded me that rewriting the whole paper in terms of quasigroups would obscure the original motivation, and we settled on a remark to the effect that the proof carried over to quasigroups, and perhaps the literature on related properties could be reexamined in that context.
@PaulBurchett, I think the paper I referred to earlier was Alayont, F., & Krzywonos, N. (2013). Rook polynomials in three and higher dimensions. Involve, a Journal of Mathematics, 6(1), 35-52.
@joro, certainly for each prime $p$ there is a cutoff $N_p$ and a period $d_p \le p^{d+2}$ for which $\forall n > N_p: f(n + d_p) \equiv f(n) \pmod{p}$. Proof: the tuples $(n \bmod p, f(n) \bmod p, \ldots, f(n+d) \bmod p)$ are the states of a deterministic finite state machine. The subtlety with trying to extend this into a proof that P-recursive functions must be composite for infinitely many $n$ is that walks in the FSM might be $\rho$-shaped rather than cycles.
I think you also want to require the variable to take all square values up to $B$, don't you? As things stand, it seems that the program x = 1 answers the question.
