Between proper and square colorings of sparse graphs

An $i$-independent set is a set of vertices whose pairwise distance is at least $i+1$. A proper coloring (resp. a square coloring) of a graph is a partition of its vertices into independent (resp. $2$-independent) sets. A packing $(1^{\ell},2^k)$-coloring of a graph is a partition of its vertices into $\ell$ independent sets and $k$ $2$-independent sets; this is an intermediate coloring between proper coloring and square coloring. We investigate classes of sparse graphs that have a proper $(\ell+1)$-coloring but no packing $(1^{\ell},2^k)$-coloring for any finite $k$. The Four Color Theorem states that every planar graph is packing $(1^4)$-colorable, and Gr\"otzsch's Theorem says every planar graph with girth at least $4$ is packing $(1^3)$-colorable. However, for every fixed $k$, we construct a planar graph with no packing $(1^{3},2^k)$-coloring and a planar graph with girth $6$ that has no packing $(1^{2},2^k)$-coloring. Moreover, for every positive integer $\ell$, we completely determine the minimum girth condition $g(\ell)$ for which every planar graph with girth at least $g(\ell)$ has a packing $(1^{\ell},2^{f(\ell)})$-coloring for some finite $f(\ell)$. Our results are actually in terms of maximum average degree. We also study the list version of packing colorings. We extend two results of Gastineau and Togni by showing every subcubic graph is both packing $(1^{1},2^6)$-choosable and packing $(1^{2},2^3)$-choosable, and our results are sharp. In addition, we strengthen Voigt's example of a planar graph that is not $4$-choosable by constructing a planar graph that is not packing $(1^{4},2^k)$-choosable for every positive integer $k$.