Let $k$ be a positive integer and $G=(V,E)$ be a $2$-connected simple graph.Suppose $v\in V(G)$ satisfy:


$(1)$there exists at least one vertex $u\in V(G)\backslash\{v\}$ such that $u$ is not adjacent with $v$;


$(2)$for any $u\in V(G)\backslash\{v\}$ such that $u$ is not adjacent with $v$,there is a $u$-$v$ path in $G$ which has an odd length $\geq2k+1$.


I want to ask is it ture that there must exists a odd cycle in $G$ whose length$\geq2k+1$.



A conjecture about odd path and odd cycle
False. As we want a counterexample, we naturally start with the Petersen graph, P. Note that for any vertex v of P and non-adjacent edge uw of P there is a Hamiltonian path from v to w that does not use the uw edge. On the other hand, there is no Hamiltonian cycle in P.


Our graph G will have 10t+15 vertices (where t is some large number) and is obtained from P by subdividing each edge with a vertex and replacing each original vertex by t vertices, connected to each other and to the subdividing vertices which the original vertex was connected to. Let v be a vertex that was not obtained by subdividing an edge and let 2k+1=9t+16 (or +17, whichever is odd). Using the above mentioned properties of P, it is easy to see that G satisfies the requirements if t is big enough.


This graph is even 3-connected and can be generalized to give even stronger counterexamples.


Note that my original construction used t=2, which is not sufficient for the above reasoning, as was noted in the comment by Dani.


