Covering triples by quadruples: An asymptotic solution

Let C(3, 4, n) be the minimum number of four-element subsets (called blocks) of an n-element set, X, such that each three-element subset of X is contained in at least one block. Let L(3, 4, n) = {top left corner} n 4{top left corner}n- 1 3{top left corner}n- 2 2{top right corner}{top right corner}{top right corner}. Schoenheim has shown that C(3, 4, n) ≥ L(3, 4, n). The construction of Steiner quadruple systems of all orders n≡2 or 4 (mod 6) by Hanani (Canad. J. Math. 12 (1960), 145-157) can be used to show that C(3, 4, n) = L(3, 4, n) for all n ≡ 2, 3, 4 or 5(od 6) and all n ≡ 1 (mod 12). The case n ≡ 7 (mod 12) is made more difficult by the fact that C(3, 4, 7) = L(3, 4, 7) + 1 and until recently no other value for C(3, 4, n) with n≡7 (mod 12) was known. In 1980 Mills showed by construction that C(3, 4, 499) = L(3, 4, 499). We use this construction and some recursive techniques to show that C(3, 4, n) = L(3, 4, n) for all n ≥ 52423. We also show that if C(3, 4, n) = L(3, 4, n) for n = 31, 43, 55 and if a certain configuration on 54 points exists then C(3, 4, n) = L(3, 4, n) for all n ≠ 7 with the possible exceptions of n = 19 and n = 67. If we assume only C(3, 4, n) = L(3, 4, n) for n = 31 and 43 we can deduce that C(3, 4, n) = L(3, 4, n) for all n ≠ 7 with the possible exceptions of n ε{lunate} {19, 55, 67, 173, 487}.