---

For the scales and exceptional set of Lemma 3.1,

$$|E|\ll \frac{x(\log_2x)^5}{\log^2x}. \tag{1}$$

Proof. It suffices to bound a union of the six classes; overlaps only help. Write $\ell=\log x$ and $u=\log\ell$.

Class 1 has at most $x/L$ elements. For class 2, Rankin's bound applied to $a_n=1_{n\le x}$ gives

$$\#\{n\le x:n\in\mathbb N_{\le R}\} \ll x^{1-1/\log R}\log R =\frac{x}{3\ell^2u},$$

because $x^{-1/\log R}=e^{-3u}=\ell^{-3}$. Class 3 has at most

$$\sum_{L<d\le\sqrt x}\frac{x}{d^2}\ll x/L$$

elements, by comparison with an integral.

For class 4, a union bound gives

$$x\sum_{\substack{d>D\\d\in\mathbb N_{\le L}}}\frac1d \ll x(\log L)D^{-1/\log L}. \tag{2}$$

To get (2), apply Rankin to $a_d=1_{d>D}/d$; its weighted supremum is at most $D^{-1/\log L}$. Since $\log D/\log L=u^2/10$, (2) is $O(xu e^{-u^2/10})$, smaller than the target for sufficiently large $x$.

For class 5 we may discard numbers already in class 4, so $d\le D$. For fixed $d,p_2$,

$$R/L\le p_2\le p_1\le \min\!\left(\frac{x}{dp_2},p_2L\right).$$

The PNT upper bound $\pi(t)\ll t/\log t$, following from Lemma 1.6, bounds the number of $p_1$ by

$$\frac{C}{\log(R/L)} \min\!\left(\frac{x}{dp_2},p_2L\right).$$

Here the upper endpoint is at least $p_2\ge R/L$ whenever a choice exists. Necessarily $p_2\le\sqrt{x/d}$. Put $T=\sqrt{x/(dL)}$. Since $d\le D$, this tends to infinity uniformly. The part with $p_2<T$ has weighted sum at most

$$\sum_{p_2<T}p_2L \le LT\pi(T)\ll\frac{x/d}{\log(x/(dL))} \ll\frac{x/d}{\log(x/(DL))}.$$

For $T\le p_2\le\sqrt{x/d}$, Lemma 1.7 and the ratio of endpoints $\sqrt L$ give

$$\sum_{T\le p_2\le\sqrt{x/d}}\frac{x}{dp_2} \ll\frac{x\log L}{d\log(x/(DL))}.$$

The exponentially small term in that lemma is bounded by a constant, and $\log L\to\infty$. Finally

$$\sum_{d\in\mathbb N_{\le L}}\frac1d\ll\log L$$

by Mertens' product. Thus class 5, apart from class 4, contributes

$$\ll\frac{x\log^2 L}{\log(R/L)\log(x/(DL))} \ll\frac{xu^3}{\ell^2}. \tag{3}$$

For class 6, fix $p_1,p_2,p_3$ and bound the number of $d$ by $x/(p_1p_2p_3)$. Enlarging the two inner prime ranges,

$$\#E_6\le x\sum_{R/L^2\le p_3\le x}\frac1{p_3} \left(\sum_{p_3\le p\le p_3L^2}\frac1p\right)^2.$$

Lemma 1.7 bounds each inner sum by $C\log(L^2)/\log(R/L^2)$. Mertens bounds the outer sum by $O(u)$. Since $\log(R/L^2)\asymp\ell/u$ and $\log L=10u$, this is $O(xu^5/\ell^2)$. Combining all six contributions proves (1). $\square$

The source's reference to $\#A_3$ during class 5 means the contribution to the exceptional set $E$. No deeper anatomy-of-integers result is needed for these bounds.

Source. Tao, published paper, published pp.802–805, Proposition 3.2. This page uses that published version.

Bears on. Problem 49.