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We'll show that the answer is $k-1$. Take the path $P$ with vertices $v_1 , v_2 , ... , v_l$ respectively , as the largest path of graph $G$. Consider the graph $G - \{v_1 , v_2 , ... , v_l\}$ and take path $P_1$ as its largest path again and construct paths $P_2 , ... , P_s$ in the same way respectively until the end of the process , thus we need to show that $s \le k-2$. Let $w_1 , w_2 , ... , w_s$ be first vertices of paths $P_1 , ... , P_s$ , we'll show that there is no two adjacent vertices between $v_1 , v_l , w_1 , ... , w_s$. Note that since $P$ is the largest path of the graph , $v_1$ and $v_l$ don't have any adjacent vertex out of $P$ and also because of the same reason , for any $1 \le i<j \le s$ , $w_i$ and $w_j$ can't be adjacent too since $P_i$ is the largest path of a subgraph without vertices of $P_1 , P_2 , ... , P_{i-1}$. It remains two prove that vertices $v_1$ and $v_l$ can't be adjacent , since $G$ is connected , there exist a vertex $w$ out of $P$ and a vertex $v_i$ ($ 2 \le i \le l-1 $) such that $v_i$ and $w$ are adjacent and if $v_1 , v_l$ are two adjacent vertices too , then the graph has a path with length $l+1$ which is a contradiction : $$w , v_i , v_{i-1} , ... , v_1 , v_l , v_{l-1} , ... ,v_{i+1}$$So while $k$ is the independence number of the graph and $v_1 , v_ l , w_1 , ... , w_s$ form an independent subgraph with $s+2$ vertices , we have $s+2 \le k$ and the path number of graph is at most $k-1$. For construction , consider a graph that contains a vertex $v$ , adjacent to $k$ other vertices $v_1 , ... , v_k$ which clearly works. So we're done.