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For sufficiently large real $x$, set

$$L=(\log x)^{10},\qquad D=\exp((\log_2x)^3),\qquad R=x^{1/(3\log_2x)}. \tag{1}$$

Let $E\subset[x]$ consist of integers satisfying at least one condition:

1. $n\le x/L$.
2. $n\in\mathbb N_{\le R}$.
3. $d^2\mid n$ for some integer $d>L$.
4. $d\mid n$ for some $d\in\mathbb N_{\le L}$ with $d>D$.
5. $n=dp_2p_1$, with $d\in\mathbb N_{\le L}$ and $R/L\le p_2\le p_1\le p_2L$.
6. $n=dp_3p_2p_1$, with $d\in\mathbb N$ and $R/L^2\le p_3\le p_2\le p_1\le p_3L^2$.

Let $A_1\subset[x]$ consist of $n=dp>x/L$ with $d\in\mathbb N_{\le L}$, $d\le D$ and $p$ prime. Let $A_2\subset[x]$ consist of $n=dps$ with $p>L$ prime, $d\in\mathbb N_{<p}$ and $s$ a prime or product of two primes, all of whose prime factors exceed $pL$. Then

$$[x]\subset A_1\cup A_2\cup E. \tag{2}$$

Proof. The scales obey

$$1<L<D^{1/\log L}<D<e^{\sqrt{\log x}}<R<x.$$

For example $\log L=10\log_2x$, $\log D=(\log_2x)^3$ and $\log R=\log x/(3\log_2x)$; each displayed inequality follows by comparing these expressions as $x\to\infty$. These comparisons also give $x/(DL)>D^A$ for each fixed $A$, eventually.

Take $n\notin E$. List its four largest prime factors, with multiplicity, as $p_1\ge p_2\ge p_3\ge p_4$, using the value one when the list ends. Then $n>x/L$ and $p_1>R$.

If $p_2\ge p_1/L$, the inequality $p_3\ge p_2/L$ would give $p_3>R/L^2$ and $p_1\le p_3L^2$, putting $n$ in class 6. Thus $p_3<p_2/L$. If $p_3\le L$, all remaining factors are at most $L$, and $p_2\ge p_1/L>R/L$ would put $n$ in class 5. Consequently $L<p_3<p_2/L$. Class 3 excludes $p_4=p_3$. Taking $p=p_3$, $s=p_1p_2$ and $d=n/(p_1p_2p_3)$ gives $A_2$.

If $L<p_2<p_1/L$, class 3 excludes $p_3=p_2$. Taking $p=p_2$, $s=p_1$ and $d=n/(p_1p_2)$ again gives $A_2$, with $p_1>pL$. Finally, if $p_2\le L$, then $d=n/p_1$ is $L$-smooth. Class 4 forces $d\le D$, giving $A_1$. These cases are exhaustive. $\square$

Source precision. The published split uses $p_2>p_1/L$ first and $L<p_2\le p_1/L$ second. The equality case does not immediately imply the strict separator in $A_2$. The repartition above proves the same lemma with its original definitions, including real $L$.

Source. Tao, published paper, published pp.800–802, scales (3.1)–(3.4) and Lemma 3.1. This page uses that published version.

Bears on. Problem 49.