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Claim. A. Sánchez-Flores, On tournaments free of large transitive subtournaments, Graphs Combin. 14 (1998), no. 2, 181--200, proves that every tournament on $54$ vertices contains a transitive subtournament on $7$ vertices, and that the tournament on $12$ vertices with no transitive subtournament on $5$ vertices and the tournament on $26$ vertices with no transitive subtournament on $6$ vertices are each unique; this is the paper's result as the zbMATH review (Zbl 0918.05058, by J. Bang-Jensen) states it, the review adding that special classes of tournaments are studied with the aid of a computer. So $R(7)\le54$ and $f(54)\ge7$, while the formula of Problem 1216 gives $\lfloor\log_254\rfloor+1=6$; the formula fails at $n=54$, where Reid and Parker's Corollary 2 gives only $6$. Stearns's doubling step $R(k+1)\le2R(k)$ turns $R(7)\le54$ into $R(k)\le54\cdot2^{k-7}$ for $k\ge7$, that is $f(n)\ge\lfloor\log_2n-\log_2(54)\rfloor+7$, the bound the site's commentary credits to this paper for $n\ge32$; it exceeds $\lfloor\log_2n\rfloor+1$ for $n$ in $[54\cdot2^j,2^{j+6})$ for each $j\ge0$.

Depends on. Nothing in this wiki; the result is the paper's own theorem.

Source. The page is dated by the issue date of the journal record (Graphs and Combinatorics 14, no. 2, 5 June 1998, per Crossref).

Acceptance. Reviewed: the site's curator, T. F. Bloom, labels the problem DISPROVED and credits Sánchez-Flores [Sa98b] in the problem's commentary with the bound $f(n)\ge\lfloor\log_2n-\log_2(54)\rfloor+7$ for $n\ge32$ (page last edited 12 April 2026); the curator is independent of the author. Refereed: Graphs and Combinatorics 14 (1998), no. 2, 181--200. The proof is not checked in this corpus.