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Let $F(n)={\prod }_{j=1}^k(a_{j}n+b_j)$ be a product of fixed pairwise nonproportional linear polynomials with positive integer leading coefficients. Assume admissibility: no prime number divides $F(n)$ for every integer $n$. There are constants $\delta ,c>0$ such that, for all large $X$, at least $cX/(\log X{)}^k$ values $1\le n\le X$ have no prime factor of $F(n)$ below $X^{\delta }$. These are almost prime values with a bounded total number of prime factors, counted with multiplicity.

A rigorous positive count needs a lower-bound sieve input, obtainable from the fundamental lemma of sieve theory. If ${\rho }_F(d)$ counts roots modulo squarefree $d$, then the polynomial root density in a sieve gives $X{\rho }_F(d)/d+O({\rho }_F(d))$. Take distribution level $D=X^{1/2}$ and sifting limit $z=X^{\delta }$. The total remainder up to $D$ is $O_k(D(\log (2D){)}^k)$, while Mertens' theorems give ${\prod }_{p<z}(1-{\rho }_F(p)/p){\asymp }_F(\log z{)}^{-k}$. Choosing $\delta$ sufficiently small makes the main-term error in the fundamental lemma of sieve theory less than half the main term, proving the lower count.^[ref]

Since $|F(n)|{\ll }_F{X}^k$, each surviving nonzero value has at most $k/\delta +O_F(1)$ prime factors, with multiplicity. The Buchstab identity organizes the removal of successive prime factors, and upper-bound sieves control removed terms; merely observing convergence of a prime tail does not establish that its subtraction leaves a positive main term.

Take $\mathcal{P}$ to contain the primes at least $5$; the exceptional local behavior at $2$ and $3$ is then harmless. Fix a sufficiently large constant $w$ and later choosewith fixed small $0<\delta <\eta$. The Buchstab identity givesThe first term is $c_{w}X+O_w(1)$ for a positive constant $c_w$, by direct counting in the finitely many permitted residue classes modulo $P_w$.

For each term in the sum, part c supplies the local factor $g(p)=3/p$ and remainders bounded by powers of $3^{\omega (d)}$. Apply the Selberg upper-bound sieve to the remaining prime conditions. Mertens theorem gives the dimension-three densityso the main terms in the Buchstab sum are bounded byThis convergent tail can be made smaller than $c_{w}X/4$ by taking $w$ large. The weighted remainder terms are $o(X)$: the estimate $k^{\omega (n)}{\ll }_{k,\epsilon }{n}^{\epsilon }$ controls the summed remainders, while part d controls uniformly the loss caused by the finite sieve level. Choosing $\delta$ sufficiently small relative to $\eta$, and then taking $X$ large, therefore givesfor some absolute $c>0$.

For every counted $m$, the distinct prime divisors of $F(m)$ are either below the fixed $w$ or at least $X^{\delta }$. The first class contains at most $\pi (w)$ primes, while $F(m)\ll X^3$ makes the second class contain at mostprimes. Since every prime divisor of $m$, $m+2$, or $m+6$ divides $F(m)$,after enlarging an absolute constant $C$. A positive proportion occurs for arbitrarily large $X$, so infinitely many such $m$ exist. This is the almost-primes from an upper-bound sieve and Buchstab identity method.

Replace $F(m)$ by $m(m+1)(m+2)$. The only change is at the locally obstructing primes: every value is divisible by $2$, and every value is divisible by $3$, so omit both primes from $\mathcal{P}$. For every $p\ge 5$, the three roots $0,-1,-2$ are distinct and again give $g(p)=3/p$ and $r_d=O(3^{\omega (d)})$. All Selberg upper-bound sieve, Buchstab identity, and large-prime-factor estimates from part e are unchanged, proving the analogous result.

For a finite set $A$ of positive integers and a set of prime numbers $\mathcal{P}$, the sifting function counts elements divisible by no prime number in $\mathcal{P}$ strictly below $z$:The strict inequality is the endpoint convention used in the Buchstab identity here. With an empty product, $P(z)=1$ and every element survives.

For a finitely supported nonnegative weighted sequence $\mathcal{A}=(a_n)$, use the same notation for $S(\mathcal{A},\mathcal{P},z)={\sum }_{\gcd (n,P(z))=1}{a}_n$. Choosing $a_n$ to be the indicator function of $A$ recovers the set version. For $A=\{1,\dots ,10\}$, all prime numbers allowed, and $z=5$, only $2$ and $3$ are removed as divisors, leaving $1,5,7$ and giving $S=3$.