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I can't comment on the question, so I will suggest an approach here. Resize the targets (partial checkerboards in R^2) by dividing by n, and stop when the union of the resizings covers the plane (or e …

It might help to start with a notion of arrangement (dominoes might overlap), note that the shuffle is an involution on the set of arrangements, and then do a tedious combinatorial argument to show it …

The 4x4 problem is similar to labelling 16 of the interior 24 edges black with some constraints: the top 3 edges must have at least one black edge, and the top 7 edges have at most 68 admissible color …

This suggests a different approach to bounding the number of configurations. E.umerate a sequence of partial configurations. Note that there are 32 choices to place a block in the upper left corner …

Notice that the product prod P ( bounded below by n) represents the order of a permutation with cycle structure given by P and sitting in S_m, where m=f(n). So considering the largest order of elemen …

Addendum: I misread the problem, not paying attention to the for all s and r. Thus the trivial answer to the second more general question trivially misses the intent. However, one can change the pro …

Even with a distinguished root and insisting that all isomorphisms respect this distinguished root, this will be a challenging enumeration. For $k \leq 2$ the problem is straightforward: The count is …

EDIT 3 These things always come to me after I post. Let's build one cage for both kings. Allocate a corner and a 2-rank by 4-file space for the cage. (One can do a vertical 4 by 2 cage also, but t …

This is more a collection of potentially useful ideas and intuitions, with no guarantee of correctness and no proof. If you take a coloring and tweak it by switching the colors on two vertices of opp …

The comments have covered the bulk of the behaviour of P(n,N) for nontrivial values of n, showing that only when n=1 should P have values exceeding 2 (or n at most 2 to get a value more than 1). A so …

There is probably a cleaner justification, but I'll give this sketch and wait for something more slick from someone else. Recall that any binary matrix of order k2^t where k is odd can be put into a …

This is a suggestion for further development, as opposed to an answer. It seems to hold much promise. Note that the n^2 upper bound can easily be shortened by 2, since any permutation not beginning …

I like intelligent brute force algorithms. While there may be more clever ones, the following is pretty simple. I will specialize it to the case of looking for $R(6,6)$. Suppose we have a list (or w …

This is essentially a question about counting nonintersecting short paths in a cubic lattice, but with a twist. (One constraint that I did not make clear below is that when to turn is already chosen …

It looks like the titled question (d=3) is not directly answered: I will hint at how to show the answer is no. At each vertex, there are two ways that the decomposition can go. I like to call them b …