In this paper (Construction 2.6 p860) the authors have built examples of connected $k$-regular graph without Hamiltonian path, but with a cut-vertex (i.e. it is not $2$-connected).

*Question:* Is there a $2$-connected $k$-regular graph without Hamiltonian path?

*Remark*: every $2$-connected $k$-regular graph with at most $3k+3$ vertices admits a Hamiltonian cycle or is one of the two exceptions, which admit a Hamiltonian path.

some$k$? or a counterexample foreach$k$? $\endgroup$ – Gerry Myerson Jun 22 '16 at 22:44