The Ramsey Number for Tree Versus Wheel with Odd Order
Statement
For even and ,
Context
Candidate 1 of the open problems stated in "The Ramsey Number for Tree Versus Wheel with Odd Order", extracted for the Scalable Mathematical Discovery run.
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Attempts
No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.
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Problem: Reconstructed statement: for finite simple graphs, is any tree on vertices, is the wheel with rim length , and for even and ,
The “odd order” title supports this convention, since even gives odd.
Result: The conjecture is false. Take , . Let be the 13-vertex tree with vertices
and edges
Then .
Construct a red/blue coloring of . Partition the vertices as
Color red all edges of the complete tripartite graph on , and color red all edges inside . Color every remaining edge blue.
First, red has no copy of . Its red components are and , so any red copy of would have to lie in . But : if the singleton part is occupied by a vertex , then must split between the two 6-vertex parts with equal sizes. For the components of , the bipartition imbalances are:
No signed sum of any listed multiset is , so no equal split is possible.
Second, blue has no . A blue wheel with hub would require a blue inside . If , then , which has no . If , then , independent. If or , then ; every cycle there uses at most five independent-side vertices and hence has length at most . Thus no blue exists.
Therefore this coloring of contains neither a red nor a blue , so
This contradicts the conjectured equality.
Citation: No external citation; the counterexample is explicit above.
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1 machine checkNo person has reviewed this attempt. 1 machine check below — a machine check is not human verification.
Machine check · not human verification
machine: correctscope Full solution as submitted; SMD novelty classification TYPE1
PASS
The proposed counterexample is mathematically sound. For , the tree has . The coloring of correctly avoids a red copy of : is too small, and the imbalance argument rules out embedding into . The blue-neighborhood analysis also correctly rules out a blue , since no possible hub has a blue in its neighborhood. Thus , contradicting the conjecture.
Novelty assessment
TYPE1
Classification rationale: The accepted solution is a genuine explicit counterexample to the finite-form conjecture, but it is a single small exceptional construction , not an exact Ramsey-number determination or an infinite family. It is useful as a correction/erratum to a specialized conjecture, but likely too small to support a standalone standard-journal paper without further extensions.
Literature check: I found no prior source containing this counterexample, the -style coloring, or a statement that Hafidh–Baskoro’s finite conjecture is false. Searches for the exact conjectural condition, , , and citation trails mostly led to partial/supporting results. Notably, Zhang–Chen–Zhang’s 2026 note says it resolves the Hafidh–Baskoro conjecture “for all large ,” so it does not cover this small- counterexample. Other recent work treats or large- regimes.
Citation: No known citation for the counterexample. Relevant literature: Hafidh–Baskoro, “The Ramsey Number for Tree Versus Wheel with Odd Order,” Bull. Malays. Math. Sci. Soc. 44 (2021), 2151–2160, DOI 10.1007/s40840-020-01055-x; Zhang–Chen–Zhang, “On tree–wheel Ramsey numbers,” Discrete Appl. Math. 383 (2026), 80–84, DOI 10.1016/j.dam.2025.12.007; Britz–Chng–Wong, Discrete Math. 348 (2025), 114461.
No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.
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