---

Let

$$\mathcal C_x=[x]\setminus\{2^kp:k\ge0,\ p\text{ prime}\}.$$

Then, for $x\ge10$,

$$M(\mathcal C_x) \le\left(\frac12+ O\!\left(\frac{(\log_2x)^5}{\log x}\right)\right)\pi(x). \tag{1}$$

Proof. Decompose $\mathcal C_x$ by Lemma 3.1. The exceptional and secondary parts retain the bounds in Proposition 3.2 and Proposition 3.3. For a primary representation $n=dp$, the ratio $\varphi(d)/d=1$ would imply $d=1$, and ratio $1/2$ would imply $d=2^k$ with $k\ge1$, by the support characterization. Both are excluded by the definition of $\mathcal C_x$. Every remaining fiber therefore has reciprocal mass at most $1/2$ by Remark 2.2.

Repeat the proof of Proposition 3.4 on this primary subfamily. Its hulls remain disjoint; the term $1/\log x$ in each fiber integral is now multiplied by at most $1/2$, while the logarithmic moment remains at most four. The same calculation bounds its size by

$$\left(\frac12+O(\log_2x/\log x)\right)x/\log x.$$

Adding the other two contributions and using the PNT gives (1) for sufficiently large $x$. A larger absolute error constant covers the bounded range $x\ge10$, exactly as in Theorem 1.1. The integer one, if present, causes no exception: for large $x$ it lies in the small exceptional class. $\square$

Source. Tao, published paper, published p.815, Section 4.3. This page uses that published version.

Bears on. Problem 49.