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For each positive integer $n$, let $P(n)$ denote the largest prime factor of $n$ (and for completeness, define $P(1) = 1$, say). Let $f(x) \in \mathbb{Z}[x]$ be an irreducible polynomial of degree at least $2$ with no fixed prime divisor (that is, for all primes $p$ there exists $n_p \in \mathbb{Z}$ such that $p \nmid f(n_p)$). Let $B$ be a positive number that we understand to be tending to infinity, and let $\delta > 0$ be a small positive number. Define $$\displaystyle \mathcal{A}_1^\delta (B) = \{n \in \mathbb{N} : n \leq B, P(f(n)) > B^{1 + \delta}\},$$ and for each $k \geq 2$, let $\mathcal{A}_k^\delta(B)$ denote the subset of $[1,B] \cap \mathbb{N}$ consisting of those $n$ for which $f(n)$ is divisible by at least $k$ distinct primes $p_1, \cdots, p_k$ such that $p_i > B^{\frac{1 + \delta}{k}}$ for $i = 1, \cdots, k$.

Clearly, for fixed $f$ and $\delta$ and any $B \in \mathbb{R}^+$, there exists a number $m = m(f,\delta, B)$ such that $$\displaystyle [1,B] \cap \mathbb{N} = \bigcup_{i=1}^m \mathcal{A}_i^\delta(B).$$ The question is, is it possible to choose, for each $f$ and $\delta$, a number $\kappa = \kappa(f,\delta)$, independent of $B$, such that $$\displaystyle \#\left( [1,B] \cap \mathbb{N} \setminus \bigcup_{i=1}^\kappa \mathcal{A}_i^\delta(B) \right) = o(B)?$$

Heath-Brown showed, in this paper (http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=74297), that for $\delta \leq 10^{-303}$ and $f(x) = x^3 + 2$, that $$\displaystyle \limsup_{B \rightarrow 0} \frac{\# \mathcal{A}_1^\delta(B)}{B} > 0.$$ This result seems to be improved in December 2014 by Irving in the following preprint: http://arxiv.org/abs/1412.0024 .

If the question can be answered at all, how large can $\delta$ be made ($\delta$ may depend on the degree of $f$) in general, or in some specific cases?

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