jmc

We can find a recursion. Let $D_n$ be the number of legal delivery sequences for $n$ houses. If a sequence ends with a delivery, we simply append one to $D_{n-1}$. If it ends in $1$ nondelivery, we append a nondelivery and a delivery to $D_{n-2}$. If it ends in $2$ nondeliveries, we append them and a delivery to $D_{n-3}$. So $D_n=D_{n-1}+D_{n-2}+D_{n-3}$. Thus, since clearly $D_1=2$, $D_2=4$, $D_3=7$, we have $D_4=13$, $D_5=24$, $D_6=44$, $D_7=81$, $D_8=149$, $D_9=274$, $D_{10}=\boxed{504}$.