Normalization of the metric f(R) field equation ($A{E}_{ab}=\frac{1}{2}{T}_{ab}$)

For a metric action $S_g=A\int \sqrt{-g}f(R)d^{4}x$ and the usual stress-energy tensor convention, covariant metric variation gives $A{E}_{ab}=T_{ab}/2$, where $E_{ab}=f^{\prime }{R}_{ab}-f{g}_{ab}/2+(g_{ab}\square -{\nabla }_a{\nabla }_b)f^{\prime }$. With $A=1/(16\pi G)$ this becomes $E_{ab}=8\pi G{T}_{ab}$; with $A=1$ it becomes $E_{ab}=T_{ab}/2$. The chosen normalization does not alter the vacuum equation.