$\displaystyle a_n(G)=\frac{t_n(G)}{(n-1)!}$ and $\displaystyle h_n(G)=\sum _{k=1}^n {n-1 \choose k-1} t_k(G)h_{n-k}(G).$ Then need to remove $\displaystyle t_k(G)$ and obtain $\displaystyle a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).$