  A graph $G$ is $(d_1,d_2,d_3)$-colorable if the vertex set $V(G)$ can be
partitioned into three subsets $V_1,V_2$ and $V_3$ such that for
$i\in\{1,2,3\}$, the induced graph $G[V_i]$ has maximum vertex-degree at most
$d_i$. So, $(0,0,0)$-colorability is exactly 3-colorability.
  The well-known Steinberg's conjecture states that every planar graph without
cycles of length 4 or 5 is 3-colorable. As this conjecture being disproved by
Cohen-Addad etc. in 2017, a similar question, whether every planar graph
without cycles of length 4 or $i$ is 3-colorable for a given $i\in
\{6,\ldots,9\}$, is gaining more and more interest. In this paper, we consider
this question for the case $i=6$ from the viewpoint of improper colorings. More
precisely, we prove that every planar graph without cycles of length 4 or 6 is
(1,0,0)-colorable, which improves on earlier results that they are
(2,0,0)-colorable and also (1,1,0)-colorable, and on the result that planar
graphs without cycles of length from 4 to 6 are (1,0,0)-colorable.
