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The Selberg upper-bound sieve replaces the indicator function of a sifted set by the square of a divisor sum. If the real Selberg sieve weights satisfy ${\lambda }_1=1$, thenWhen $F(n)$ has no forbidden prime factors, the divisor sum is exactly one; otherwise its square is still nonnegative. Taking the weights to vanish for $d>z$ controls the error in summing this majorant. The main term is a positive quadratic form that can be diagonalized and minimized explicitly. This optimization, together with a bound for the remainder, is the central idea of the Selberg sieve.

For twin primes, take $F(n)=n(n+2)$ and let $\rho (d)$ be the number of its roots modulo a squarefree integer $d$. The Chinese remainder theorem makes $\rho$ a multiplicative arithmetic function, with $\rho (2)=1$ and $\rho (p)=2$ for odd primes. Thus the polynomial root density in a sieve is $g(d)=\rho (d)/d$, andSumming the majorant giveswhere $Q(\lambda )={\sum }_{d,e\le z}{\lambda }_d{\lambda }_{e}g([d,e])$, with all indices squarefree. Here $\rho ([d,e])\le \rho (d)\rho (e)$ controls the error.

We carry out the Selberg sieve diagonalization. PutFor squarefree integers $d,e$, multiplicativity givesThe identity behind the first equality is ${\prod }_{p\mid (d,e)}[1+h(p)]=1/g((d,e))$. Möbius inversion gives ${\lambda }_1={\sum }_r\mu (r)y_r=1$. The Cauchy-Schwarz inequality therefore impliesDefine $G(z)={\sum }_{r\le z,r\text{ squarefree}}1/h(r)$. Equality is attained at $y_r=\mu (r)/(h(r)G(z))$, giving $\min Q=1/G(z)$. Inverting again, the optimizing Selberg sieve weights areIn particular ${\lambda }_1=1$. Since $G_d(z/d)\le G(z)$,The last inequality holds separately at $p=2$, $p=3$, and $p\ge 5$. Together with $\rho (d)\le d$, this crude bound already makes the remainder $O(z^6)$, because ${\sum }_{d\le z}{d}^2=O(z^3)$.

We also need a lower bound for $G(z)$; we prove it rather than presuppose a sieve asymptotic. For odd squarefree integers $r$,where $\omega (r)$ counts distinct prime factors. Put $y=\sqrt{z}$ and $L(y)={\sum }_{a\le y,a\text{ odd}}1/a=\frac{1}{2}\log y+O(1)$. The total weight $1/(ab)$ of odd pairs $a,b\le y$ is $L(y{)}^2$. Exclude pairs for which $a$ has a square prime factor, $b$ has a square prime factor, or $a,b$ have a common prime factor. For each of these three categories the total weight is at most $L(y{)}^2{\sum }_{p\text{ odd}}{p}^{-2}$: for example, the weighted count of odd multiples of $p^2$ is $p^{-2}L(y/p^2)\le p^{-2}L(y)$, and the common-factor category uses $p^{-1}L(y/p)$ for each variable. Moreover,Hence the remaining odd coprime squarefree pairs have weight at least $L(y{)}^2/6$. Their products $r=ab\le z$ are squarefree, and $2^{\omega (r)}$ counts all ordered coprime factorizations of $r$. This proves the elementary lower bound for the twin-prime sieve denominator:

If $p>z$ and both $p,p+2$ are primes, $F(p)$ survives the sieve. The at most $z$ smaller values are the only exceptions. Thus, with $T(x)=\#\{p\le x:p,p+2\text{ prime}\}$ and $z=x^{1/10}$, the twin-prime upper bound from a quadratic sieve becomesFinally partial summation givesThe integral over $[3,\infty )$ is bounded by a constant times ${\int }_3^{\infty }dt/(t(\log t{)}^2)<\infty$. The nonnegative partial sums are therefore bounded and increasing. Thus the reciprocal series over the smaller members of twin-prime pairs converges, a form of Brun's theorem. The upper bound does not assert that infinitely many twin primes exist.

Let $A(x)$ count primes $p\le x$ for which $p+2$ is also prime. We prove $A(x)\ll x/{\log }^{2}x$ using a quadratic Selberg sieve, then apply Abel summation.

For $F(n)=n(n+2)$, each odd prime has exactly two roots modulo that prime. The Chinese remainder theorem gives the polynomial root density in a sievefor odd squarefree integers $d$. Put $R=x^{1/8}$ and choose the optimizing real Selberg sieve weights ${\lambda }_d$, supported on these $d\le R$, with ${\lambda }_1=1$. If $p,p+2$ are prime and $p>R$, their weight $({\sum }_{d\mid F(p)}{\lambda }_d{)}^2$ is one. It is nonnegative for every other integer. Expanding the square therefore givesThe Selberg sieve diagonalization makes the main quadratic form $1/J(R)$, whereThe optimal Selberg weights have modulus at most one. Since ${\rho }_F([d,e])\le [d,e]\le R^2$, the remainder is at most $O(R^4)$.

For completeness, the lower bound on $J$ is where the second logarithm enters. Restrict to primes $3\le p\le y=R^{1/8}$ and put $k(p)=2/(p-2)$. The full product $W={\prod }_{3\le p\le y}(1+k(p))$ is $\asymp {\log }^{2}y$, since its logarithm is $2{\sum }_{3\le p\le y}1/p+O(1)$ and the Mertens second theorem applies. Give the associated squarefree divisor $d$ probability $k(d)/W$. Prime inclusion probabilities are $2/p$, so the Mertens first theorem givesfor large $R$. The Markov inequality shows that at least half the weight has $d\le R$. This proves the truncated Euler-product lower bound $J(R)\gg W\gg {\log }^{2}R$. Consequently the twin-prime upper bound from a quadratic sieve isFinally, Abel summation givesThe tail is dominated by ${\int }_3^{\infty }dt/(t{\log }^{2}t)<\infty$. Thus reciprocal-sum convergence from a counting bound proves convergence of the first reciprocal sum. The second is smaller termwise. We have proved Brun's theorem:Counting each twin prime just once changes at most a finite repetition and gives the same conclusion.

Letand let $W$ be the product of the finitely many primes at most $r$ or dividing some nonzero difference $h_i-h_j$. Sieve with the primes $p\nmid W$. For such a prime, the congruence $F(n)\equiv 0(\mathrm{mod}p)$ has exactly $r$ distinct roots. By the Chinese remainder theorem, the root-counting function $\rho (d)$ is multiplicative on squarefree $d$ coprime to $W$, andThus the polynomial root density in a sieve applies with

Take $z=x^{1/4}$ and $D=z$. For squarefree $d$ coprime to $W$,The supplied mean-value estimate, partial summation, and removal of square factors using $r^{\omega (n)}{\ll }_{r,ϵ}{n}^{ϵ}$ giveThe error in the Selberg upper-bound sieve isConsequently

If every $n+h_i$ is prime and all of them exceed $z$, then $F(n)$ has no prime divisor $p<z$ with $p\nmid W$, apart from a fixed finite set of divisibility cases absorbed into the implied constant. The cases with some $n+h_i\le z$ contribute $O_r(z)$. Therefore

Put $F(m)=m(m+2)(m+6)$ and let $\rho (d)$ be the number of roots of $F$ modulo $d$. For squarefree $d$, the Chinese remainder theorem givesIndeed, for $p\ge 5$ the three roots $0,-2,-6$ are distinct. Counting $m\le X$ in each of the $\rho (d)$ residue classes givesConsequently a suitable sieve distribution isThis is the polynomial root density in a sieve calculation.