Question:
what can be said about the existence of $2k$ regular graphs, $1\lt k$ that have a $1$-factor and a $2$-factor?
Provided their existence, what is/are the smallest for $k$?
The graphs must be simple, i.e. they must not neither have self-loops nor multiple edges.
All simple non-empty regular graphs of even degree have a two factor, see here . So you are just asking when they have a 1-factor. In addition to having an even number of vertices, the conditions are known explicitly but aren't simple, see here . The conditions usually hold and it has been proved that almost all non-empty regular graphs with an even number of vertices have a 1-factor, in the sense of the proportion going to 1 as the number of vertices increases.
Take the graph on $2k+2$ vertices $x_0, \ldots, x_k, y_0, \ldots, y_k$ with an edge between any pair of distinct vertices except $(x_i, y_i)$ for $0\le i \le k$. There are plenty of $2$-factors, e.g. the Hamiltonian cycle $x_0 \to \ldots \to x_k \to y_0 \to \ldots \to y_k \to x_0$. Removing all the edges on this Hamiltonian cycle also gives a $1$-factor.
This is clearly the smallest number of vertices you can have since a any $2k$-regular simple graph has at least $2k+1$ vertices and the existence of a $1$-factor forces an even number of vertices.
