Erdős Problem #1182
Status Open · erdosproblems.comLLM status Open unverified
Erdős problems openaigpt 6 astra ultra 10% cataloguedOpenOpenStatement
Let $R(H,G)$ denote the two-colour Ramsey number. For each $n$, define $f(n)$ as the greatest edge count of a connected $n$-vertex graph $G$ satisfying $R(K_3,G)=2n-1$. Define $F(n)$ as the greatest threshold such that every connected $n$-vertex graph with at most $F(n)$ edges satisfies this equality. Determine the asymptotic sizes of $f(n)$ and $F(n)$. The original questions include whether $F(n)/n\to\infty$.
Restatement from Math-6236-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6236
Erdős Problem 1182 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Let $R(H,G)$ denote the two-colour Ramsey number. For each $n$, define $f(n)$ as the greatest edge count of a connected $n$-vertex graph $G$ satisfying $R(K_3,G)=2n-1$. Define $F(n)$ as the greatest threshold such that every connected $n$-vertex graph with at most $F(n)$ edges satisfies this equality. Determine the asymptotic sizes of $f(n)$ and $F(n)$. The original questions include whether $F(n)/n\to\infty$.CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.The official remarks already answer the final subquestion negatively via Brandt's bound F(n)<=84n; the estimation problem remains open. Definitions are intended in the nontrivial large-n regime.
ATTACK PLAN AND WORK
Every connected $n$-vertex graph has Ramsey number at least $2n-1$ against a triangle. On $2n-2$ vertices, make the red graph complete bipartite with two parts of size $n-1$. It has no red triangle, while the blue graph consists of two disjoint cliques, neither large enough to contain a connected $n$-vertex graph.For every tree $T$ on $n$ vertices the lower bound is attained. Consider any coloring on $2n-1$ vertices with no red triangle. If a vertex has at least $n$ red neighbors, those neighbors form a blue clique, which contains $T$. Otherwise every vertex has red degree at most $n-1$, hence blue degree at least $n-1$. A graph of minimum degree at least $n-1$ contains every $n$-vertex tree: root the tree and embed vertices in parent-before-child order; whenever a new child is needed, its already embedded parent has at most $n-2$ other occupied vertices to avoid and at least $n-1$ neighbors. Thus \[R(K_3,T)=2n-1.\tag{1}\]
Every connected graph with $n-1$ edges is a tree, so $1$ gives the baseline $F(n)\ge n-1$ and $f(n)\ge n-1$. These are independently proved elementary benchmarks, not a claim of a new tree Ramsey theorem. The source's cited linear upper bound for $F$ answers its superlinear subquestion negatively; that external upper-bound proof is not reconstructed here.