3-Vertices with fewest 2-neighbors in plane graphs with no long paths of 2-vertices

Let $g(k,t)$ be the minimum integer such that every plane graph with girth g at least $g(k,t)$, minimum degree $\delta =2$ and no $(k+1)$-paths consisting of vertices of degree 2, where $k\ge 1$, has a 3-vertex with at least t neighbors of degree 2, where $1\le t\le 3$.In 2015, Jendrol' and Maceková proved $g(1,1)\le 7$. Later on, Hudák et al. established $g(1,3)=10$, Jendrol', Maceková, Montassier, and Soták proved $g(1,1)\ge 7$, $g(1,2)=8$ and $g(2,2)\ge 11$, and we recently proved that $g(2,2)=11$ and $g(2,3)=14$.Thus $g(k,t)$ is already known for $k=1$ and all t. In this paper, we prove that $g(k,1)=3k+4$, $g(k,2)=3k+5$, and $g(k,3)=3k+8$ whenever $k\ge 2$.