Claims

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1970_10_01_reid_parker: Reid and Parker's Theorem 4 (J. Combinatorial Theory 1970): every tournament on 14 vertices contains a transitive subtournament on 5 vertices, so f(14) = 5 while floor(log_2 14) + 1 = 4; the answer is no.

1994_06_01_neumann_lara: Neumann-Lara (Graphs Combin. 1994) reproves that for k >= 5 every tournament of order at least 7 * 2^(k-4) contains a transitive k-subtournament; at k = 5, f(14) >= 5 > 4, so the formula fails.

1994_06_01_sanchez_flores: Sánchez-Flores (Graphs Combin. 1994): every tournament on 55 vertices contains a transitive subtournament on 7 vertices, so f(55) >= 7 while floor(log_2 55) + 1 = 6; the formula fails.

1998_06_05_sanchez_flores: Sánchez-Flores (Graphs Combin. 1998): every tournament on 54 vertices contains a transitive subtournament on 7 vertices, so f(54) >= 7 while floor(log_2 54) + 1 = 6; the formula fails.

2020_11_02_neiman_mackey_heule: Neiman, Mackey and Heule (Graphs Combin. 2022), computer-assisted: every tournament on 47 vertices contains a transitive 7-subtournament, so f(n) >= 7 > floor(log_2 n) + 1 for 47 <= n <= 63.

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