Let $s_n=a_1+\dots+a_n$ so that $s_{n+1}=s_n+P(s_n)$ where $P(x)$ is the greatest prime factor of $x$, i.e. $P(s_n)=a_{n+1}$. One can now easily show by induction that $s_n$ moves from $p_{k-1}p_k$ to $p_kp_{k+1}$ in $(p_{k+1}-p_{k-1})$ steps of size $p_k$. Since $s_4=2 \cdot 3$, we then have $s_{p_{k-1}+p_k-1}=p_{k-1}p_k$. So $s_{99}=47 \cdot 53$ and hence $\boxed{a_{100}=53}$.