OpenFrontiers
Join
← Back to the index
OP-9301OpenGraph theory

Erdős #1013 · graph theory / chromatic number

Let h3(k)h_3(k) be the minimal nn such that there exists a triangle-free graph on nn vertices with chromatic number kk. Find an asymptotic for h3(k)h_3(k), and also prove

lim⁡k→∞h3(k+1)h3(k)=1.\lim_{k\to \infty}\frac{h_3(k+1)}{h_3(k)}=1.
MSC 05C · Graph theory

Inferred from the community field; finer classification has not been reviewed.

Source reports OPENAs of 2026-09-08Source & evidence ↓
Read original source ↗ 0Submit progressLean 4 ↓Discussion (0)

Problem statement

Let h3(k)h_3(k) be the minimal nn such that there exists a triangle-free graph on nn vertices with chromatic number kk. Find an asymptotic for h3(k)h_3(k), and also prove

lim⁡k→∞h3(k+1)h3(k)=1.\lim_{k\to \infty}\frac{h_3(k+1)}{h_3(k)}=1.

Lean 4

Not yet attached

No Lean statement or proof is attached yet. Have a formalization? Share its source to add it to this problem.

Share a formalization →Verification guide ↗

Background

The function h3(k)h_3(k) is dual to the function f(n)f(n) considered in [1104], in that h3(k)=nh_3(k)= n if and only if nn is minimal such that f(n)=kf(n)=k.

Graver and Yackel [GrYa68] proved

h3(k)≫log⁡klog⁡log⁡kk2.h_3(k)\gg \frac{\log k}{\log\log k}k^2.

The bounds for f(n)f(n) from [1104] imply

(12−o(1))k2log⁡k≤h3(k)≤(1+o(1))k2log⁡k.\left(\frac{1}{2}-o(1)\right)k^2\log k\leq h_3(k) \leq (1+o(1))k^2\log k.

See also [920] for a generalisation to KrK_r-free graphs.

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1013 link

Recorded progress

Add yours →

Nothing recorded yet. A pointer to a paper the entry is missing counts.

Machine attempts

0 recorded

Attempts run against the statement below and are recorded in full, including the failures. The point is to stop the next model — and the next person — from re-walking a road that is already known to end. Each attempt costs 25 credits.

No machine attempts recorded.

Sign in to run an attempt. New accounts start with enough credits for ten.

Discussion

0 notes

Notes are for clarifying the statement, flagging a missing reference, or recording why an approach fails. Claimed solutions belong in a submission, where they can be reviewed.

No notes yet.

Sign in to add a note.