Determining if an input undirected graph is Hamiltonian, i.e., if it has a cycle that visits every vertex exactly once, is one of the most famous NP-complete problems. We consider the following generalization of Hamiltonian cycles: for a fixed set $S$ of natural numbers, we want to visit each vertex of a graph $G$ exactly once and ensure that any two consecutive vertices can be joined in $k$ hops for some choice of $k \in S$. Formally, an $S$-Hamiltonian cycle is a permutation $(v_0,\ldots,v_{n-1})$ of the vertices of $G$ such that, for $0 \leq i \leq n-1$, there exists a walk between $v_i$ and $v_{i+1 \bmod n}$ whose length is in $S$. (We do not impose any constraints on how many times vertices can be visited as intermediate vertices of walks.) Of course Hamiltonian cycles in the standard sense correspond to $S=\{1\}$. We study the $S$-Hamiltonian cycle problem of deciding whether an input graph $G$ has an $S$-Hamiltonian cycle. Our goal is to determine the complexity of this problem depending on the fixed set $S$. It is already known that the problem remains NP-complete for $S=\{1,2\}$, whereas it is trivial for $S=\{1,2,3\}$ because any connected graph contains a $\{1,2,3\}$-Hamiltonian cycle. Our work classifies the complexity of this problem for most kinds of sets $S$, with the key new results being the following: we have NP-completeness for $S = \{2\}$ and for $S = \{2, 4\}$, but tractability for $S = \{1, 2, 4\}$, for $S = \{2, 4, 6\}$, for any superset of these two tractable cases, and for $S$ the infinite set of all odd integers. The remaining open cases are the non-singleton finite sets of odd integers, in particular $S = \{1, 3\}$. Beyond cycles, we also discuss the complexity of finding $S$-Hamiltonian paths, and show that our problems are all tractable on graphs of bounded cliquewidth.