I decided to look at the hexagonal version of
https://oeis.org/A277433 Martin Gardner's minimal no-3-in-a-line problem, all slopes version.
Not shown: the unique 5-point solution for the side-3 hexagon. I've turned
this into a Martin Gardner style puzzle.
Remove three cells each from two opposite corners of a 5×5 grid.
On this grid, place five counters such that adding one more counter on any vacant square will produce three in a line. The solution is unique (barring rotations and reflections)
🟨🟨🟨🟫🟫🟨🟨🟨🟨🟫🟨🟨🟨🟨🟨🟫🟨🟨🟨🟨🟫🟫🟨🟨🟨 Here are my best solutions for higher orders.