3.2. 孪生素数个数上界的改良

我们考虑集合

$$\mathcal{A}=\{n(n-h):n\le x\},$$

则很明显对于所有 $z\ge 0$ 均有:

$${\pi }_h(x)\le z+S(\mathcal{A},\mathcal{P},z).$$(3.2.1)

而结合 $\mathcal{A}$ 本身的性质, 不难发现:

$$X=x,g(p)=\{\begin{array}{ll}1/p& p|h\\ 2/p& p\nmid h\end{array},|r(d)|\le 2^{\omega (d)}.$$

现在套用定理 3.4.1, 可知 $\xi =z$ 时:

$$S(\mathcal{A},\mathcal{P},z)<\frac{x}{G}+\sum _{\begin{array}{c}d|P(z)\\ d<z^2\end{array}}{6}^{\omega (d)}.$$(3.2.2)

对于余项, 利用 Cauchy–Schwarz 不等式知:

$$\begin{array}{rl}\sum _{\begin{array}{c}d|P(z)\\ d<z^2\end{array}}{6}^{\omega (d)}& =\sum _{\begin{array}{c}d|P(z)\\ d<z^2\end{array}}\frac{6^{\omega (d)}}{\sqrt{d}}\sqrt{d}\le {\left(\sum _{d|P(z)}\frac{36^{\omega (d)}}{d}\right)}^{1/2}{\left(\sum _{d<z^2}d\right)}^{1/2}\\ & <z^2\prod _{p<z}{\left(1+\frac{36}{p}\right)}^{1/2}<z^2\exp \left\{\sum _{p<z^2}\frac{18}{p}\right\}\ll z^2{\log }^{18}z.\end{array}$$

对于主项, 当 $h$ 固定、$s\to 0^{+}$ 时:

$$\begin{array}{rl}\sum _{d\ge 1}\frac{{\mu }^2(d)h(d)}{d^s}& =\prod _p\left(1+\frac{h(p)}{p^s}\right)\\ & =\prod _{p|h}\left(1+\frac{1}{p^s(p-1)}\right)\prod _{p\nmid h}\left(1+\frac{2}{p^s(p-2)}\right)\\ & =(1+2^{-s})\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{1+p^{-s}(p-1{)}^{-1}}{1+2p^{-s}(p-2{)}^{-2}}\prod _{p>2}\left(1+\frac{2}{p^s(p-2)}\right)\\ & \sim 2\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-2}{p-1}\prod _{p>2}\left(1+\frac{2}{p^s(p-2)}\right).\end{array}$$

对于右侧的乘积, 由于:

$$\sum _p\log \left(1+\frac{2}{p^s(p-2)}\right)=\sum _p\frac{2}{p^s}+O\left(\sum _p\frac{1+p^s}{p^{2s+2}}\right),$$(3.2.3)

所以有:

$$\begin{array}{rl}\prod _{p>2}\left(1+\frac{2}{p^s(p-2)}\right)& \sim \prod _{p>2}\left(1+\frac{2}{p-2}\right){\left(1-\frac{1}{p}\right)}^2\prod _{p>2}{\left(1-\frac{1}{p^{s+1}}\right)}^{-2}\\ & =(1-2^{-s-1}{)}^2\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}\prod _p{\left(1-\frac{1}{p^{s+1}}\right)}^{-2}\\ & \sim \frac{1}{4}\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}{\zeta }^2(s+1).\end{array}$$

因此当 $u_r$ 满足:

$$\sum _{r\ge 1}\frac{u_r}{r^s}=\frac{1}{{\zeta }^2(s+1)}\sum _{d\ge 1}\frac{{\mu }^2(d)h(d)}{d^s}$$(3.2.4)

时根据 (3.2.3) 可知, (3.2.4) 在 $s>-1/2$ 均绝对收敛. 因此当 $\tau (n)$ 表示 $n$ 的正因子个数时总有:

$$\begin{array}{rl}G& =\sum _{d<z}{\mu }^2(d)h(d)=\sum _{r<z}{u}_r\sum _{n<z/r}\frac{\tau (n)}{n}\\ & =\sum _{r<z}{u}_r\left[\sum _{k<z/r}\frac{1}{k}\sum _{m<z/rk}\frac{1}{m}\right]\\ & =\sum _{r<z}{u}_r\left\{\sum _{k<z/r}\frac{1}{k}\left[\log \frac{z}{rk}+O(1)\right]\right\}\\ & =\sum _{r<z}{u}_r\left[\log \frac{z}{r}\sum _{k<z/r}\frac{1}{k}-\sum _{k<z/r}\frac{\log m}{m}+O(1)\right]\\ & =\sum _{r<z}{u}_r\left[\frac{1}{2}{\log }^2\frac{z}{r}+O(\log z)\right]\\ & =\frac{1}{2}{\log }^{2}z\sum _{r<z}{u}_r+O\left\{\log z\sum _{r<z}|u_r|{\log }^{2}r\right\}\\ & =\frac{1}{2}{\log }^{2}z\sum _{r\ge 1}{u}_r+O(\log z)\\ & =\frac{1}{4}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-2}{p-1}\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}{\log }^{2}z+O(\log z),\end{array}$$

因此将 $z=x^{1/2}{\log }^{-11}$ 代入到 (3.2.2) 和 (3.2.1) 中, 就有:

定理 3.2.0.1. 对于所有固定的正偶数 $h$, 总有: $${\pi }_h(x)\le \frac{16x}{{\log }^{2}x}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-1}{p-2}\prod _{p>2}\left(1-\frac{1}{(p-1{)}^2}\right)+O\left(\frac{x\log \log x}{{\log }^{3}x}\right).$$