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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.