for $a,b,c\geq 0$ and $0\leq m \leq a+b+c+1$, $$\label{bcalc} \omega_{a,b} \circ_m \partial^c \alpha^i = \lambda_{a,b,c,m} \partial^{a+b+c+1-m} \alpha^i$$ where $$\lambda_{a,b,c,m}  = (-1)^b \frac{(b+c+1)!}{(b+c+1-m)!} + (-1)^a \frac{(a+c+1)!}{(a+c+1-m)!}.$$ It follows that for $m\leq a+b+c+1$ we have $$\label{opeformula} \omega_{a,b}\circ_m \omega_{c,d} = \lambda_{a,b,c,m} \omega_{a+b+c+1-m,d} + \lambda_{a,b,d,m}\omega_{c,a+b+d+1-m}.$$ In fact, there is a somewhat more economical set of strong generators for $\cH(n)^{O(n)}$. For each $m\geq 0$, let $A_m$ denote the vector space spanned by $\{\omega_{a,b}|~ a+b = m\}$, which is homogeneous of weight $m+2$. Clearly $\text{dim}(A_{2m}) = m+1 = \text{dim}(A_{2m+1})$ for $m\geq 0$. Moreover, $\partial(A_m)\subset A_{m+1}$, and we have $$\label{deca} \text{dim} \big(A_{2m} / \partial(A_{2m-1})\big) = 1,\ \ \ \ \ \ \text{dim} \big(A_{2m+1} / \partial(A_{2m})\big) = 0.$$ For $m\geq 0$, define $$\label{defofj} j^{2m} = \omega_{0,2m},$$ which is clearly not a total derivative. Hence $A_{2m}$ has a decomposition $$\label{decompofa} A_{2m} = \partial (A_{2m-1})\oplus \bra j^{2m}\ket =  \partial^2 (A_{2m-2})\oplus \bra j^{2m}\ket ,$$ where $\bra j^{2m}\ket$ is the linear span of $j^{2m}$. Similarly, $A_{2m+1}$ has a decomposition $$\label{decompofai} A_{2m+1} = \partial^2(A_{2m-1})\oplus \bra \partial j^{2m}\ket =  \partial^3 (A_{2m-2})\oplus \bra \partial j^{2m}\ket.$$ It is easy to see that $\{\partial^{2i} j^{2m-2i}|~ 0\leq i\leq m\}$ and $\{\partial^{2i+1} j^{2m-2i}|\ 0\leq i\leq m\}$ are bases of $A_{2m}$ and $A_{2m+1}$, respectively. Hence each $\omega_{a,b}\in A_{2m}$ and $\omega_{c,d}\in 