For n=2 the upper bound is 6.82… according your last formula. Is it OK?

]]>But this bound is asymptotically worse than the known upper bound (Larman, Rogers, 1972)?

]]>Oh, actually, in trying to formalize the odd bound, I got a slightly better and cleaner bound: 2*(2+sqrt(n))^(n-1).

]]>Great! I’ve found a better bound: ((2+sqrt(n-1))^n)/(sqrt(n)/2) for floor(2+sqrt(n-1)) even, and (2+sqrt(n))^n/(sqrt(n)/2) for floor(2+sqrt(n-1)) odd. So I guess the latter as a universal bound; I’m working on making it work better for that one, and I can save 1/2*(floor(2+sqrt(n))-1)^n colors so far, but that converges to a small fraction of the total number of colors, 1/e^(sqrt(n)), which itself converges to 0, so I don’t think it’s relevant yet.

]]>As far as I know, there is no such bound.

]]>The key difference is that in the Moser spindle case we are dealing with the addition of elements that do not belong to the previous field.

In case (2,3), the element belongs to the cyclotomic field . In fact, we can get the same set of vertices by adding the degrees of divided by 3. Here .

Therefore a similar coloring, if it exists, should be based on homomorphism

and here the finite ring contains elements.

Perhaps there is a simpler way.

]]>For any case of a generalized spindle we can take pi/3 . The problem is the size of the 4-chromatic subgraphs, and, informally speaking, their density.

The reasoning on this wiki page is not too complicated, the meaning is clear, but I don’t have the proper technique and can make silly mistakes.

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