Tratnik et al.’s nanotube resonance-graph conjecture

The conjecture asserts that for every matchable nanotube NN, its resonance graph R6(N)R_6(N) is disconnected. Here R6(N)R_6(N) is the graph whose vertices are the perfect matchings of NN, with two perfect matchings MM and MM' adjacent when there is a hexagonal face HH such that MM=E(H)M\mathbin{\triangle}M'=E(H).

Equivalent formulations 1

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

  1. Connectivity criterion for nanotube resonance graphs

    For every matchable nanotube NN, the resonance graph R6(N)R_6(N) is connected if and only if every nice cycle of NN is an I-cycle or bounds a hexagonal system.

    source: Liang, Lingmei, Zhang, and Heping, “The resonance graphs of coronoid systems and nanotubes”

Sources & referencesView supporting material

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A new paper says the conjecture is false by giving a nanotube with a connected resonance graph, although the result has not yet been independently verified.

Tratnik et al.’s conjecture asserts that the resonance graph R6(N)R_6(N) of every matchable nanotube NN is disconnected. A new paper directly challenges this assertion while confirming it for the narrower class of elementary nanotubes.

August 2026 counterexample

Liang, Lingmei, Zhang, and Heping report a matchable nanotube N1N_1 with connected R6(N1)R_6(N_1) and state that infinitely many such examples exist. They also give a criterion: R6(N)R_6(N) is connected exactly when every nice cycle is an I-cycle or bounds a hexagonal system; for elementary nanotubes, equality of flows across every cut segment characterizes resonance-graph components. The paper is a direct arXiv report, with no independent verification or referee assessment found.

Current status (as of August 2026): The universal conjecture is claimed false by an explicit connected resonance graph, while the elementary-nanotube case is reported to remain true; independent confirmation is pending.

Sources

Solutions 0

No solutions have been posted yet.