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Claim. A. Sánchez-Flores, On tournaments and their largest transitive subtournaments, Graphs Combin. 10 (1994), no. 2--4, 367--376, proves that every tournament of order $55$ contains a transitive subtournament of order $7$, and that the tournament of order $27$ with no transitive subtournament of order $6$ is unique; this is the paper's result as the zbMATH review (Zbl 0811.05029, by B. Alspach) states it. So the directed Ramsey number satisfies $R(7)\le55$ and $f(55)\ge7$, while the formula of Problem 1216 gives $\lfloor\log_255\rfloor+1=6$; the formula fails at $n=55$, a value where Reid and Parker's Corollary 2 gives only $\lfloor\log_2(16\cdot55/7)\rfloor=6$. Stearns's doubling step $R(k+1)\le2R(k)$ turns $R(7)\le55$ into $R(k)\le55\cdot2^{k-7}$ for $k\ge7$, that is $f(n)\ge\lfloor\log_2n-\log_2(55)\rfloor+7$ for $n\ge55$, the bound the site's commentary credits to this paper; it exceeds $\lfloor\log_2n\rfloor+1$ for $n$ in $[55\cdot2^j,2^{j+6})$ for each $j\ge0$. The zbMATH review places $R(7)\le55$ in this 1994 paper, as the site does; Nagy's introduction, which attributes the bound in the form $F(n)\ge\lfloor\log_2(n/55)\rfloor+7$ to the 1998 paper, is in error on this point.

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

Source. The page is dated by the issue month of the journal record (Graphs and Combinatorics 10, no. 2--4, June 1994, per Crossref); the day in the page name is a placeholder.

Acceptance. Reviewed: the site's curator, T. F. Bloom, labels the problem DISPROVED and credits Sánchez-Flores [Sa94] in the problem's commentary with the bound $f(n)\ge\lfloor\log_2n-\log_2(55)\rfloor+7$ for $n\ge55$ (page last edited 12 April 2026); the curator is independent of the author. Refereed: Graphs and Combinatorics 10 (1994), no. 2--4, 367--376. The proof is not checked in this corpus.