 Twin-prime upper bound from a quadratic sieve

For $F(n)=n(n+2)$, every odd prime has two forbidden residue classes. Use Selberg sieve weights supported on odd squarefree integers $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J={\sum }_{d\le R}{\prod }_{p\mid d}2/(p-2)$. The truncated Euler-product lower bound applied to primes up to $R^{1/8}$ gives $J\gg {\log }^{2}R$: the mean logarithmic divisor size is $2{\sum }_{3\le p\le R^{1/8}}(\log p)/p<\frac{1}{2}\log R$, by the Mertens first theorem, and the full product is $\asymp {\log }^{2}R$, by the Mertens second theorem. The optimal Selberg weights have modulus at most one, so the remainder is at most $O({\sum }_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.