Hitting a hub from two four-cycles

Take two four-cycles, connect corresponding vertices, and connect each inner vertex to an absorbing hub. For the ordinary random walk on a graph, symmetry reduces mean remaining times to outer and inner types, satisfying $O=1+(2O+I)/3$ and $I=1+(O+2I)/4$. Thus $O=10$, $I=7$. For the non-backtracking random walk, directed-edge types outer-outer, outer-inner, inner-outer and inner-inner give remaining means $13/2,9/2,15/2,21/4$. Starting at an outer vertex, the first-step distribution yields total mean $41/6$.