Let $A(x)=\{n(n+2):1\le n\le x\}$. For squarefree $d\mid P(z)$, suppose \begin{align*} |A_d|=\frac{\omega(d)}{d}x+O(\omega(d)), \end{align*} where $\omega$ is multiplicative, $\omega(2)=1$, and $\omega(p)=2$ for odd primes. There is an absolute constant $c>0$ such that, for $z=x^{1/u}$ with $2\le u\le c\log\log x$, an even Brun truncation depth can be chosen so that \begin{align*} S(A(x),P,z)\ll x\prod_{p<z}\left(1-\frac{\omega(p)}{p}\right). \end{align*} The implicit constant is absolute.