$$a_0=0,a_1=1,a_2=2,a_3=4,a_4=5,\dotsc.$$

$$\varlimsup_{n\rightarrow\infty}d_n= \infty.$$

$\begin{cases}x+1\equiv p_1\pmod{p_1^2},\\ x+2\equiv p_2\pmod{p_2^2},\\ \cdots\\x+n\equiv p_n\pmod{p_n^2}.\end{cases}$

$S:=\{a_0,a_1,a_2,a_3,\dotsc\}.$

$$S$$ 的密率是 $$0$$. 事实上, 我们有精确得多的结果.

$$\lim_{n\rightarrow\infty}\frac{A(n)\sqrt{\log n}}n=1.$$

$$L_{1}(s)=\sum_{n\,\text{odd}}(-1)^\frac{n-1}2 n^{-s}=\prod_{p\in\Pi_1}\dfrac{1}{1-p^{-s}}\prod_{p\in\Pi_3}\dfrac1{1+p^{-s}}.$$

$$L_{0}(s)=\sum_{n\,\text{odd}}n^{-s}=\prod_{p\in\Pi_1}\dfrac1{1-p^{-s}}\prod_{p\in\Pi_3}\dfrac1{1-p^{-s}}=(1-2^{-s})\zeta(s),$$

$$\log L_0+\log L_1=2\sum_{p\in\Pi_1}p^{-s}+O(1),$$

$$\log L_0-\log L_1=2\sum_{p\in\Pi_3}p^{-s}+O(1).$$

$$L_1$$ 是交换级数, 因而收敛, 有界, 并且当 $$s\geqslant1$$ 时, 其上下界与 $$s$$ 无关. 于是

$$\sum_{p\in\Pi_1}p^{-s}=\frac12\log\left[(1-2^{-s})\zeta (s)\right]+O(1),$$

$$\sum_{p\in\Pi_3}p^{-s}=\frac12\log\left[(1-2^{-s})\zeta (s)\right]+O(1).$$

$$\prod_{p\in\Pi_1}\frac1{1-p^{-s}}=\sqrt{(1-2^{-s})\zeta (s)}(1+O(1)),$$

$$\prod_{p\in\Pi_3}\frac1{1-p^{-s}}=\sqrt{(1-2^{-s})\zeta (s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_n^{-s}=\frac1{1-2^{-s}}\prod_{p\in\Pi_1}\frac1{1-p^{-s}}\prod_{p\in\Pi_3}\frac1{1-p^{-2s}}=\sqrt{\zeta (s)\zeta (2s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_n^{-s}=\sqrt{\zeta(s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_{n}^{-s}=\sum_{n=1}^\infty\chi(n)n^{-s},$$

$$\chi(n)=\begin{cases}\,1,\quad\,n\in S\\0,\quad \text{otherwise}\end{cases}$$

$A(x)=\sum_{n\leqslant x}\chi(n),$

$$\sqrt{\zeta (s)}+O(1)=s\int_{0}^\infty A(x)x^{-(s+1)}dx.$$

$$\frac1{\sqrt{s-1}}+O(1)=\int_{0}^\infty A(x)x^{-(s+1)}dx.$$

$$A(x)\sim\frac{x}{\sqrt{\log x}}.$$

$\{\frac{s_1}{s_2}|s_1,s_2\in \Bbb S\}$

$g(x)=a_n, h(x)=a_{n+1}(a_n\leqslant x<a_{n+1},n=0,1,2,\dotsc.),$

$\dfrac{\dfrac{g(np)}{h(nq)}}{\dfrac{h(np)}{g(nq)}}<\dfrac{\dfrac{g(np)}{g(nq)}}{\dfrac pq}<\dfrac{\dfrac{h(np)}{g(nq)}}{\dfrac{g(np)}{h(nq)}},$

$\lim_{n\rightarrow\infty}\frac{g(np)}{g(nq)}=\frac pq.$

$$\sum_{n=1}^\infty\frac1{a_n}$$

$\sum_{p\equiv b\pmod a\atop (a,b)=1}\frac 1p=\infty$

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