Selberg upper-bound sieve

Suppose $|{\mathcal{A}}_d|=Xg(d)+r_d$ for $d\mid P(z)$. If ${\lambda }_1=1$ and the real Selberg weights ${\lambda }_d$ vanish unless $d\mid P(z)$ and $d\le D$, thenOptimizing the main quadratic form gives $X/G(D,z)$, whereup to the harmless replacement of $D$ by $D^2$ under the alternative convention that $D$ denotes the level of the least common multiples.

For an integer polynomial $F$ and squarefree $d$, let ${\rho }_F(d)$ count the roots of $F$ modulo $d$. The Chinese remainder theorem makes ${\rho }_F$ multiplicative, andThus $g(d)={\rho }_F(d)/d$ and $r_d=O({\rho }_F(d))$ define a sieve distribution.

For a product of finitely many admissible linear forms, first omit a fixed finite set of locally obstructing primes. Apply the Buchstab identity at a large fixed $w$, and bound each removed term by the Selberg upper-bound sieve. The convergent tail ${\sum }_{p\ge w}1/(p(\log p{)}^k)$ leaves a positive proportion with no prime divisor in $[w,X^{\delta })$. Their distinct prime divisors below $w$ are finite in number, while those above $X^{\delta }$ number at most $O(1/\delta )$, producing infinitely many almost-prime values.

If $f$ is smooth and supported on $[-1,1]$ and $g(t)={\int }_{\mathbb{R}}{e}^{x}f(x)e(-tx)dx$, Fourier inversion givesExpanding the square of the resulting Möbius-weighted divisor sum turns its mean value into an Euler product controlled by zeta functions near their pole at one.

For $Y\ge X\ge 2$, a smooth Selberg divisor weight of level $D=X^{1/10}$ equals one on every prime in $(Y,Y+X]$. Its second moment can be expressed through an Euler product $H$ whose zeta-factor bound contributes $O(1/\log D)$ after integration against rapidly decreasing Fourier transforms. Consequently