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Math-6158-A1 · revision r1 · on Math-6158

Reused writeup on Math-6158 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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Contents
  1. Erdős Problem 1104 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 1104 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1104

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1104.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1104.tex.

FORMAL RESTATEMENT. A (simple) graph is triangle-free if it has no $3$-cycle. Let $f(n)$ be the maximum possible chromatic number among all triangle-free graphs on $n$ vertices: \[f(n):=\max\{\chi(G): |V(G)|=n,\ G\ \text{triangle-free}\}. \] Similarly, let $g(m)$ be the maximum possible chromatic number among all triangle-free graphs with $m$ edges: \[g(m):=\max\{\chi(G): |E(G)|=m,\ G\ \text{triangle-free}\}. \] Question: estimate $f(n)$ (and $g(m)$).

QUICK LITERATURE/CONTEXT CHECK (from the problem file only). The text records that bounds for the Ramsey number $R(3,k)$ imply $f(n)\asymp (n/\log n)^{1/2}$ and gives the best known constant-factor bounds \[(1-o(1))(n/\log n)^{1/2}\le f(n)\le (2+o(1))(n/\log n)^{1/2}. \] It also records bounds for $g(m)$: an upper bound $g(m)\le (3^{5/3}+o(1))(m/(\log m)^2)^{1/3}$ and a lower bound $g(m)\gg (m/(\log m)^2)^{1/3}$. I do not reprove these deep results.

ATTACK PLAN.

  • Provide a fully rigorous (but weaker) construction showing $f(n)\to\infty$ using the Mycielski construction.
  • Prove (elementary) structural lemmas about Mycielski graphs: triangle-freeness, chromatic number increment, and explicit vertex counts under iteration.
  • Reality check: compute small instances of Mycielski graphs and verify triangle-freeness for small parameters.

WORK.

Definition (Mycielski construction). Given a graph $G=(V,E)$ with $V=\{v_1,\dots,v_n\}$, define $\mu(G)$ to be the graph with vertex set \[V(\mu(G)):=V\ \cup\ \{u_1,\dots,u_n\}\ \cup\ \{w\} \] and edges:

  • all original edges $v_iv_j\in E$ remain,
  • for each original edge $v_iv_j\in E$, add edges $u_iv_j$ and $u_jv_i$,
  • connect $w$ to every $u_i$.
(No edges are added among the $u_i$.)

Lemma 1 (Mycielski preserves triangle-freeness). If $G$ is triangle-free, then $\mu(G)$ is triangle-free.

Proof. Suppose for contradiction that $\mu(G)$ contains a triangle. If the triangle uses $w$, then it would have the form $w-u_i-x$ with $x$ adjacent to both $w$ and $u_i$. But $w$ is adjacent only to $u$-vertices, and there are no edges among $u$-vertices, so no triangle can contain $w$.

Thus any triangle lies in $V\cup\{u_1,\dots,u_n\}$. There are no edges among the $u_i$, so the triangle contains at most one $u$-vertex.

If it contains no $u$-vertex, then it is a triangle entirely in $V$, contradicting that $G$ is triangle-free. So the triangle contains exactly one $u$-vertex, say $u_i$, and two original vertices $v_j,v_k$. Edges $u_i v_j$ and $u_i v_k$ exist in $\mu(G)$ only if $v_i v_j\in E(G)$ and $v_i v_k\in E(G)$. The third edge $v_jv_k$ must also be in $E(G)$ (it is an edge among original vertices). Therefore $v_i,v_j,v_k$ form a triangle in $G$, again a contradiction. Hence $\mu(G)$ is triangle-free. $\square$

Lemma 2 (Mycielski increases chromatic number by $1$). For every graph $G$, \[\chi(\mu(G))=\chi(G)+1. \]

Proof. Let $k:=\chi(G)$.

Upper bound $\chi(\mu(G))\le k+1$. Fix a proper $k$-coloring $c:V\to\{1,\dots,k\}$ of $G$. Color each $v_i$ with $c(v_i)$, color each $u_i$ with $c(v_i)$, and color $w$ with a new color $k+1$. Edges among the $v_i$ are properly colored by assumption. There are no $u_i u_j$ edges. If $u_i v_j$ is an edge in $\mu(G)$, then $v_i v_j$ was an edge of $G$, so $c(v_i)\ne c(v_j)$; hence $u_i$ and $v_j$ get different colors. Finally, $w$ has color $k+1$ different from every $u_i$. Thus this is a proper $(k+1)$-coloring of $\mu(G)$, so $\chi(\mu(G))\le k+1$.

Lower bound $\chi(\mu(G))\ge k+1$. Suppose for contradiction that $\mu(G)$ has a proper $k$-coloring $c$ with colors $\{1,\dots,k\}$. Let $\alpha:=c(w)$ be the color of $w$. Because $w$ is adjacent to every $u_i$, we have $c(u_i)\ne \alpha$ for all $i$. Let \[X:=\{v_i\in V: c(v_i)=\alpha\} \] be the set of original vertices that share $w$'s color. Since $c$ is proper and edges among $V$ in $\mu(G)$ are exactly the edges of $G$, the set $X$ is an independent set in $G$.

Define a new coloring $\tilde c$ of the original vertex set $V$ by \[\tilde c(v_i):=\begin{cases} c(u_i), & v_i\in X,\\ c(v_i), & v_i\notin X. \end{cases} \] This coloring uses only colors in $\{1,\dots,k\}\setminus\{\alpha\}$, because if $v_i\in X$ then $\tilde c(v_i)=c(u_i)\ne\alpha$ and if $v_i\notin X$ then $c(v_i)\ne\alpha$ by definition. Hence $\tilde c$ uses at most $k-1$ colors. We check that $\tilde c$ is proper on every edge $v_iv_j\in E(G)$.

If neither endpoint lies in $X$, then $\tilde c(v_i)=c(v_i)$ and $\tilde c(v_j)=c(v_j)$, and $c(v_i)\ne c(v_j)$ because $c$ is proper. If exactly one endpoint lies in $X$, say $v_i\in X$ and $v_j\notin X$, then $\tilde c(v_i)=c(u_i)$ and $\tilde c(v_j)=c(v_j)$. Because $v_iv_j\in E(G)$, the edge $u_i v_j$ is present in $\mu(G)$, so properness of $c$ gives $c(u_i)\ne c(v_j)$. Therefore $\tilde c(v_i)\ne \tilde c(v_j)$. If both endpoints lie in $X$, then there is no edge $v_iv_j$ because $X$ is independent. Thus $\tilde c$ is a proper coloring of $G$ using at most $k-1$ colors, contradicting $\chi(G)=k$. This contradiction shows that $\chi(\mu(G))\ge k+1$.

Combining the two bounds yields $\chi(\mu(G))=k+1$. $\square$

Lemma 3 (iterating Mycielski from $C_5$ gives an explicit lower bound for $f(n)$). Let $M_3:=C_5$ and define $M_{k+1}:=\mu(M_k)$ for $k\ge 3$. Then for all $k\ge 3$:

  • $M_k$ is triangle-free,
  • $\chi(M_k)=k$,
  • $|V(M_k)|=3\cdot 2^{k-2}-1$.
Consequently, for all $n\ge 5$, \[f(n)\ge 2+\lfloor \log_2\left(\frac{n+1}{3}\right)\rfloor. \]

Proof. We argue by induction on $k$. For $k=3$, $M_3=C_5$ is triangle-free and $3$-chromatic, and $|V(M_3)|=5=3\cdot 2^{1}-1$. Assume the claims for $M_k$. Then $M_{k+1}=\mu(M_k)$ is triangle-free by Lemma 1 and has chromatic number $\chi(M_k)+1=k+1$ by Lemma 2. Also, by construction $|V(\mu(G))|=2|V(G)|+1$, so $n_{k+1}:=|V(M_{k+1})|$ satisfies the recurrence $n_{k+1}=2n_k+1$ with $n_3=5$. Solving gives $n_k=3\cdot 2^{k-2}-1$.

For the bound on $f(n)$: if $n\ge n_k$, we can take $M_k$ and (if needed) add isolated vertices to reach exactly $n$ vertices; adding isolated vertices does not create triangles and does not decrease chromatic number. Thus $f(n)\ge \chi(M_k)=k$ whenever $n\ge n_k$. The inequality $n\ge 3\cdot 2^{k-2}-1$ is equivalent to $2^{k-2}\le (n+1)/3$, i.e. $k\le 2+\log_2((n+1)/3)$. Taking the largest integer $k$ satisfying this yields the displayed lower bound. $\square$

FAST REALITY CHECK (small Mycielski graphs). I generated the standard Mycielski graphs $M_k$ for $k=3,4,5,6,7$ (starting from $C_5$) and counted triangles. The results were: \[ \begin{array}{c|c|c} k & |V(M_k)| & \#\text{triangles in }M_k\\\hline 3 & 5 & 0\\ 4 & 11 & 0\\ 5 & 23 & 0\\ 6 & 47 & 0\\ 7 & 95 & 0 \end{array} \] This matches Lemma 1 and the vertex-count formula in Lemma 3.

VERIFICATION.

  • Lemma 1: enumerated all possible triangle placements and reduced to a triangle in $G$.
  • Lemma 2: the recoloring argument relies crucially on the fact that $X=\{v_i:c(v_i)=c(w)\}$ is independent in $G$ (because $c$ is a proper coloring on $V$).
  • Lemma 3: checked the recurrence $|V(\mu(G))|=2|V(G)|+1$ and solved it explicitly.

FINAL: UNRESOLVED. (i) Strongest proved partial result here. I proved from scratch that the Mycielski construction preserves triangle-freeness (Lemma 1) and increases chromatic number by $1$ (Lemma 2). Iterating from $C_5$ gives explicit triangle-free graphs with chromatic number $k$ on $3\cdot 2^{k-2}-1$ vertices, hence $f(n)\ge 2+\lfloor\log_2((n+1)/3)\rfloor$ (Lemma 3).

(ii) First gap (crisp). I did not provide a self-contained derivation of the correct order of growth $f(n)\asymp \sqrt{n/\log n}$ (or any nontrivial upper bound of that form); the missing inputs are the deep Ramsey/independence-number estimates cited in the problem text.

(iii) Top 3 next moves. 1. Work out in detail the implication "$R(3,k)\asymp k^2/\log k$ $\Rightarrow$ $f(n)\asymp \sqrt{n/\log n}$" in both directions, clearly isolating which Ramsey bounds are used where. 2. For $g(m)$, combine explicit constructions (e.g. sparse high-chromatic triangle-free graphs) with edge bounds to obtain elementary lower bounds, then compare to the $(m/(\log m)^2)^{1/3}$ scale cited. 3. Perform small-$n$ computations for triangle-free graphs (e.g. via known catalogues) to sanity-check $f(n)$ for $n\le 20$ and see the onset of growth.

(iv) Minimal counterexample structure. A counterexample to the quoted asymptotic scale would require a triangle-free graph on $n$ vertices whose chromatic number deviates substantially from the Ramsey-predicted regime, i.e. either an unexpectedly large chromatic number without creating large independent sets, or an unexpectedly small chromatic number despite the absence of large independent sets; both scenarios are governed by extremal behavior of triangle-free graphs and Ramsey theory.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.