引言
虽然已经翻译了GPY论文的两节,但是最主要的内容并没有翻译,也就是用命题1 和命题2 去证明最重要的定理1 和定理2 ,因此这也是不得不细品的一个环节.
原文的链接可以见:https://www.jstor.org/stable/25662161 .
当然,前面两节的拙劣翻译也可见我的前一篇博客.后续我将看一下命题1和命题2的证明过程,然后更新张益唐的想法与思路,而这一部分应该会更加简略一些.然后补充一下陈定理的博客,现在应该会轻松很多了吧.最后就是Maynard和Polymath的论文部分,当然也是前者为主.
命题和定理
以下是我们需要的命题.而其证明过程我后边再简单看一下好了.
命题1. 令H = H 1 ∪ H 2 \mathcal{H} = \mathcal{H}_1 \cup \mathcal{H}_2 H = H 1 ∪ H 2 ∣ H i ∣ = k i |\mathcal{H_i}| = k_i ∣ H i ∣ = k i r = ∣ H 1 ∩ H 2 ∣ r = |\mathcal{H_1} \cap \mathcal{H_2}| r = ∣ H 1 ∩ H 2 ∣ R ≪ N 1 2 ( log N ) − 4 M R \ll N^{\frac{1}{2}} (\log N)^{-4M} R ≪ N 2 1 ( log N ) − 4 M C > 0 C > 0 C > 0 h ≤ R C h \le R^C h ≤ R C R , N → ∞ R,N \to \infty R , N → ∞
∑ n ≤ N Λ R ( n ; H 1 , ℓ 1 ) Λ R ( n ; H 2 , ℓ 2 ) = ( ℓ 1 + ℓ 2 ℓ 1 ) ( log R ) r + ℓ 1 + ℓ 2 ( r + ℓ 1 + ℓ 2 ) ! ( S ( H ) + o M ( 1 ) ) N . ( 2.14 ) \sum_{n \le N} \Lambda_R(n; \mathcal{H}_1, \ell_1)\Lambda_R(n; \mathcal{H}_2, \ell_2) = \dbinom{\ell_1 + \ell_2}{\ell_1} \frac{(\log R)^{r + \ell_1 + \ell_2}}{(r + \ell_1 + \ell_2)!} (\mathfrak{S}(\mathcal{H}) + o_M(1))N.\quad (2.14)
n ≤ N ∑ Λ R ( n ; H 1 , ℓ 1 ) Λ R ( n ; H 2 , ℓ 2 ) = ( ℓ 1 ℓ 1 + ℓ 2 ) ( r + ℓ 1 + ℓ 2 )! ( log R ) r + ℓ 1 + ℓ 2 ( S ( H ) + o M ( 1 )) N . ( 2.14 )
命题2. 令H = H 1 ∪ H 2 \mathcal{H} = \mathcal{H}_1 \cup \mathcal{H}_2 H = H 1 ∪ H 2 ∣ H i ∣ = k i |\mathcal{H_i}| = k_i ∣ H i ∣ = k i r = ∣ H 1 ∩ H 2 ∣ r = |\mathcal{H_1} \cap \mathcal{H_2}| r = ∣ H 1 ∩ H 2 ∣ 1 ≤ h 0 ≤ h 1 \le h_0 \le h 1 ≤ h 0 ≤ h H 0 = H ∪ h 0 \mathcal{H}^0 = \mathcal{H} \cup {h_0} H 0 = H ∪ h 0 B ( M ) B(M) B ( M ) R ≪ M N 1 4 ( log N ) − B ( M ) R \ll_M N^{\frac{1}{4}(\log N)^{-B(M)}} R ≪ M N 4 1 ( l o g N ) − B ( M ) h ≤ R h \le R h ≤ R
∑ n ≤ N Λ R ( n ; H 1 , ℓ 1 ) Λ R ( n ; H 2 , ℓ 2 ) θ ( n + h 0 ) = { ( ℓ 1 + ℓ 2 ℓ 1 ) ( log R ) r + ℓ 1 + ℓ 2 ( r + ℓ 1 + ℓ 2 ) ! ( S ( H 0 ) + o M ( 1 ) ) N if h 0 ∉ H , ( ℓ 1 + ℓ 2 + 1 ℓ 1 + 1 ) ( log R ) r + ℓ 1 + ℓ 2 + 1 ( r + ℓ 1 + ℓ 2 + 1 ) ! ( S ( H 0 ) + o M ( 1 ) ) N if h 0 ∈ H 1 and h 0 ∉ H 2 , ( ℓ 1 + ℓ 2 + 2 ℓ 1 + 1 ) ( log R ) r + ℓ 1 + ℓ 2 + 1 ( r + ℓ 1 + ℓ 2 + 1 ) ! ( S ( H 0 ) + o M ( 1 ) ) N if h 0 ∈ H 1 ∩ H 2 . ( 2.15 ) \begin{array}{l}\sum_{n \le N} \Lambda_R(n; \mathcal{H}_1, \ell_1)\Lambda_R(n; \mathcal{H}_2, \ell_2)\theta(n+h_0) \\
=\left\{ \begin{array}{ll}
\dbinom{\ell_1+\ell_2}{\ell_1} \dfrac{(\log R)^{r + \ell_1 + \ell_2}}{(r+\ell_1+\ell_2)!} (\mathfrak{S}(\mathcal{H}^0)+o_M(1))N & \text{if } h_0 \not\in \mathcal{H}, \\
\dbinom{\ell_1+\ell_2+1}{\ell_1+1} \dfrac{(\log R)^{r + \ell_1 + \ell_2+1}}{(r+\ell_1+\ell_2+1)!} (\mathfrak{S}(\mathcal{H}^0)+o_M(1))N & \text{if } h_0 \in \mathcal{H}_1 \text{ and } h_0 \not\in \mathcal{H}_2, \\
\dbinom{\ell_1+\ell_2+2}{\ell_1+1} \dfrac{(\log R)^{r + \ell_1 + \ell_2+1}}{(r+\ell_1+\ell_2+1)!} (\mathfrak{S}(\mathcal{H}^0)+o_M(1))N & \text{if } h_0 \in \mathcal{H}_1 \cap \mathcal{H}_2. \\
\end{array} \right.
\end{array}\quad (2.15) ∑ n ≤ N Λ R ( n ; H 1 , ℓ 1 ) Λ R ( n ; H 2 , ℓ 2 ) θ ( n + h 0 ) = ⎩ ⎨ ⎧ ( ℓ 1 ℓ 1 + ℓ 2 ) ( r + ℓ 1 + ℓ 2 )! ( log R ) r + ℓ 1 + ℓ 2 ( S ( H 0 ) + o M ( 1 )) N ( ℓ 1 + 1 ℓ 1 + ℓ 2 + 1 ) ( r + ℓ 1 + ℓ 2 + 1 )! ( log R ) r + ℓ 1 + ℓ 2 + 1 ( S ( H 0 ) + o M ( 1 )) N ( ℓ 1 + 1 ℓ 1 + ℓ 2 + 2 ) ( r + ℓ 1 + ℓ 2 + 1 )! ( log R ) r + ℓ 1 + ℓ 2 + 1 ( S ( H 0 ) + o M ( 1 )) N if h 0 ∈ H , if h 0 ∈ H 1 and h 0 ∈ H 2 , if h 0 ∈ H 1 ∩ H 2 . ( 2.15 )
其中,θ \theta θ
然后就是我们将要证明的两个定理.
定理1. 假设素数的分布水平ϑ > 1 / 2 \vartheta>1/2 ϑ > 1/2 C ( ϑ ) C(\vartheta) C ( ϑ ) ϑ \vartheta ϑ k ≥ C ( ϑ ) k \ge C(\vartheta) k ≥ C ( ϑ ) k k k ϑ ≥ 0.971 \vartheta \ge 0.971 ϑ ≥ 0.971 k ≥ 6 k \ge 6 k ≥ 6
定理2. 我们有
Δ 1 : = lim inf n → ∞ p n + 1 − p n log p n = 0. ( 1.8 ) \Delta_1 := \liminf_{n \to \infty}\frac{p_{n+1}-p_n}{\log p_n} = 0.\quad (1.8)
Δ 1 := n → ∞ lim inf log p n p n + 1 − p n = 0. ( 1.8 )
以上就是本文最最关键的成果.并且其证明方式也与Maynard的证明思路很相似.
定理的证明
定理1前半部分的证明
当ℓ ≥ 0 \ell \ge 0 ℓ ≥ 0 H k = { h 1 , h 2 , ⋯ , h k } \mathcal{H}_k = \{ h_1, h_2, \cdots, h_k \} H k = { h 1 , h 2 , ⋯ , h k } 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h ≤ R 1 \le h_1, h_2, \cdots, h_k \le h \le R 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h ≤ R R ≪ N 1 2 ( log N ) − B ( M ) R \ll N^{\frac{1}{2}}(\log N)^{-B(M)} R ≪ N 2 1 ( log N ) − B ( M ) R , N → ∞ R,N \to \infty R , N → ∞
∑ n ≤ N Λ R ( n ; H k , ℓ ) 2 ∼ 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) S ( H k ) N ( log R ) k + 2 ℓ . ( 3.1 ) \sum_{n \le N}\Lambda_R(n; \mathcal{H}_k, \ell)^2 \sim \frac{1}{(k+2\ell)!} \dbinom{2\ell}{\ell} \mathfrak{S}(\mathcal{H}_k) N(\log R)^{k+2\ell}. \quad (3.1)
n ≤ N ∑ Λ R ( n ; H k , ℓ ) 2 ∼ ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ . ( 3.1 )
而对于任意h i ∈ H k h_i \in \mathcal{H}_k h i ∈ H k R ≪ N ϑ 2 − ϵ R \ll N^{\frac{\vartheta}{2} - \epsilon} R ≪ N 2 ϑ − ϵ R , N → ∞ R,N \to \infty R , N → ∞
∑ n ≤ N Λ R ( n , H k , ℓ ) 2 θ ( n + h i ) ∼ 2 ( k + 2 ℓ + 1 ) ! ( 2 ℓ + 1 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ + 1 . ( 3.2 ) \sum_{n \le N}\Lambda_R(n, \mathcal{H}_k, \ell)^2 \theta(n + h_i) \sim \frac{2}{(k+2\ell+1)!} \dbinom{2\ell+1}{\ell} \mathfrak{S}(\mathcal{H}_k) N(\log R)^{k+2\ell+1}. \quad (3.2)
n ≤ N ∑ Λ R ( n , H k , ℓ ) 2 θ ( n + h i ) ∼ ( k + 2 ℓ + 1 )! 2 ( ℓ 2 ℓ + 1 ) S ( H k ) N ( log R ) k + 2 ℓ + 1 . ( 3.2 )
于是取R = N ϑ 2 − ϵ R = N^{\frac{\vartheta}{2} - \epsilon} R = N 2 ϑ − ϵ
S : = ∑ n = N + 1 2 N ( ∑ i = 1 k θ ( n + h i ) − log 3 N ) Λ R ( n ; H k , ℓ ) 2 ∼ k 2 ( k + 2 ℓ + 1 ) ! ( 2 ℓ + 1 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ + 1 − log 3 N 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) S ( H k ) N ( log R ) k + 2 ℓ ∼ ( 2 k k + 2 ℓ + 1 2 ℓ + 1 ℓ + 1 log R − log 3 N ) 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) S ( H k ) N ( log R ) k + 2 ℓ . ( 3.3 ) \begin{array}{ll}
\mathcal{S} & := \displaystyle\sum_{n=N+1}^{2N} \left( \displaystyle\sum_{i=1}^k \theta(n+h_i)-\log 3N \right)\Lambda_R(n;\mathcal{H}_k,\ell)^2 \\
\\
& \sim k\dfrac{2}{(k+2\ell+1)!}\dbinom{2\ell+1}{\ell}\mathfrak{S}(\mathcal{H}_k)N(\log R)^{k+2\ell+1} \\
\\
& \quad -\log 3N \dfrac{1}{(k+2\ell)!}\dbinom{2\ell}{\ell}\mathfrak{S}(\mathcal{H}_k)N(\log R)^{k+2\ell} \\
\\
& \sim \left( \dfrac{2k}{k+2\ell+1} \dfrac{2\ell+1}{\ell+1} \log R - \log 3N \right)\dfrac{1}{(k+2\ell)!}\dbinom{2\ell}{\ell}\mathfrak{S}(\mathcal{H}_k)N(\log R)^{k+2\ell}.
\end{array}\quad (3.3) S := n = N + 1 ∑ 2 N ( i = 1 ∑ k θ ( n + h i ) − log 3 N ) Λ R ( n ; H k , ℓ ) 2 ∼ k ( k + 2 ℓ + 1 )! 2 ( ℓ 2 ℓ + 1 ) S ( H k ) N ( log R ) k + 2 ℓ + 1 − log 3 N ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ ∼ ( k + 2 ℓ + 1 2 k ℓ + 1 2 ℓ + 1 log R − log 3 N ) ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ . ( 3.3 )
注意,其中∑ i = 1 k θ ( n + h i ) − log 3 N \sum_{i=1}^k \theta(n+h_i) - \log 3N ∑ i = 1 k θ ( n + h i ) − log 3 N S > 0 S > 0 S > 0 H k \mathcal{H}_k H k R = N ϑ 2 − ϵ R = N^{\frac{\vartheta}{2} - \epsilon} R = N 2 ϑ − ϵ
k k + 2 ℓ + 1 2 ℓ + 1 ℓ + 1 ϑ > 1. ( 3.4 ) \dfrac{k}{k+2\ell+1}\dfrac{2\ell+1}{\ell+1}\vartheta > 1.\quad (3.4)
k + 2 ℓ + 1 k ℓ + 1 2 ℓ + 1 ϑ > 1. ( 3.4 )
其中k , ℓ → ∞ k,\ell \to \infty k , ℓ → ∞ ℓ = o ( k ) \ell = o(k) ℓ = o ( k ) 2 ϑ 2\vartheta 2 ϑ ϑ > 1 / 2 \vartheta > 1/2 ϑ > 1/2 证完定理1的前半部分 了.
并且通过(3.4),我们便可以得到不同的ϑ \vartheta ϑ k k k ℓ \ell ℓ h ( k ) h(k) h ( k ) k k k
ϑ \vartheta ϑ k k k ℓ \ell ℓ h ( k ) h(k) h ( k )
1 1 1 7 7 7 1 1 1 20 20 20
0.95 0.95 0.95 8 8 8 1 1 1 26 26 26
0.90 0.90 0.90 9 9 9 1 1 1 30 30 30
0.85 0.85 0.85 11 11 11 1 1 1 36 36 36
0.80 0.80 0.80 16 16 16 1 1 1 60 60 60
0.75 0.75 0.75 21 21 21 2 2 2 84 84 84
0.70 0.70 0.70 31 31 31 2 2 2 140 140 140
0.65 0.65 0.65 51 51 51 3 3 3 252 252 252
0.60 0.60 0.60 111 111 111 5 5 5 634 634 634
0.55 0.55 0.55 421 421 421 10 10 10 2956 ∗ 2956^* 295 6 ∗
其中2956 ∗ 2956^* 295 6 ∗
我们发现,当ϑ = 1 \vartheta = 1 ϑ = 1 k = 7 k = 7 k = 7 lim inf n → ∞ p n + 1 − p n ≤ 20 \displaystyle\liminf_{n \to \infty} p_{n+1}-p_n \le 20 n → ∞ lim inf p n + 1 − p n ≤ 20 k = 6 k = 6 k = 6
定理2的证明
而证明定理2的时候,我们需要对S \mathcal{S} S
S ~ : = ∑ n = N + 1 2 N ( ∑ 1 ≤ h 0 ≤ h θ ( n + h 0 ) − v log 3 N ) ∑ 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct Λ R ( n ; H k , ℓ ) 2 , ( 3.5 ) \begin{array}{l}
\widetilde{\mathcal{S}} := \displaystyle\sum_{n=N+1}^{2N} \left( \displaystyle\sum_{1 \le h_0 \le h} \theta(n+h_0) - v \log 3N \right) \displaystyle\sum_{\substack{1 \le h_1, h_2, \cdots, h_k \le h \\ \text{distinct}}} \Lambda_R(n; \mathcal{H}_k, \ell)^2,
\end{array} \quad (3.5) S := n = N + 1 ∑ 2 N ( 1 ≤ h 0 ≤ h ∑ θ ( n + h 0 ) − v log 3 N ) 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct ∑ Λ R ( n ; H k , ℓ ) 2 , ( 3.5 )
这时候,我们便需要使用到命题2,然后再利用Gallagher[9]中的一个结果,也就是当h → ∞ h \to \infty h → ∞
∑ 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct S ( H k ) ∼ h k . ( 3.7 ) \sum_{\substack{1 \le h_1, h_2, \cdots, h_k \le h \\ \text{distinct}}} \mathfrak{S}(\mathcal{H}_k) \sim h^k. \quad (3.7)
1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct ∑ S ( H k ) ∼ h k . ( 3.7 )
仍然取R = N ϑ 2 − ϵ R = N^{\frac{\vartheta}{2} - \epsilon} R = N 2 ϑ − ϵ
S ~ ∼ ∑ 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct ( k 2 ( k + 2 ℓ + 1 ) ! ( 2 ℓ + 1 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ + 1 + ∑ 1 ≤ h 0 ≤ h h 0 ≠ h i , 1 ≤ i ≤ k 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) S ( H k ∪ { h 0 } ) N ( log R ) k + 2 ℓ − v log 3 N 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) S ( H k ) N ( log R ) k + 2 ℓ ) ∼ ( 2 k k + 2 ℓ + 1 2 ℓ + 1 ℓ + 1 log R + h − v log 3 N ) 1 ( k + 2 ℓ ) ! ( 2 ℓ ℓ ) N h k ( log R ) k + 2 ℓ . ( 3.8 ) \begin{array}{ll}
\widetilde{\mathcal{S}} & \sim \displaystyle\sum_{\substack{1 \le h_1, h_2, \cdots, h_k \le h \\ \text{distinct}}} \left( k \dfrac{2}{(k+2\ell+1)!} \dbinom{2\ell+1}{\ell} \mathfrak{S}(\mathcal{H}_k) N (\log R)^{k + 2\ell + 1} \right. \\
\\
& \quad + \displaystyle\sum_{\substack{1 \le h_0 \le h \\ h_0 \neq h_i, 1 \le i \le k}} \dfrac{1}{(k+2\ell)!} \dbinom{2\ell}{\ell} \mathfrak{S}(\mathcal{H}_k \cup \{h_0\}) N (\log R)^{k + 2\ell} \\
\\
& \quad - \left. v \log 3N \dfrac{1}{(k+2\ell)!} \dbinom{2\ell}{\ell} \mathfrak{S}(\mathcal{H}_k) N (\log R)^{k + 2\ell} \right) \\
\\
& \sim \left( \dfrac{2k}{k+2\ell+1} \dfrac{2\ell+1}{\ell+1} \log R + h - v \log 3N \right) \dfrac{1}{(k+2\ell)!}\dbinom{2\ell}{\ell} N h^k (\log R)^{k+2\ell}.
\end{array}\quad (3.8) S ∼ 1 ≤ h 1 , h 2 , ⋯ , h k ≤ h distinct ∑ ( k ( k + 2 ℓ + 1 )! 2 ( ℓ 2 ℓ + 1 ) S ( H k ) N ( log R ) k + 2 ℓ + 1 + 1 ≤ h 0 ≤ h h 0 = h i , 1 ≤ i ≤ k ∑ ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) S ( H k ∪ { h 0 }) N ( log R ) k + 2 ℓ − v log 3 N ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) S ( H k ) N ( log R ) k + 2 ℓ ) ∼ ( k + 2 ℓ + 1 2 k ℓ + 1 2 ℓ + 1 log R + h − v log 3 N ) ( k + 2 ℓ )! 1 ( ℓ 2 ℓ ) N h k ( log R ) k + 2 ℓ . ( 3.8 )
其中∑ 1 ≤ h 0 ≤ h θ ( n + h 0 ) − v log 3 N > 0 \displaystyle\sum_{1 \le h_0 \le h} \theta(n+h_0) - v \log 3N > 0 1 ≤ h 0 ≤ h ∑ θ ( n + h 0 ) − v log 3 N > 0 ( n , n + h ] (n, n+h] ( n , n + h ] v + 1 v + 1 v + 1
h > ( v − 2 k k + 2 ℓ + 1 2 ℓ + 1 ℓ + 1 ( ϑ 2 − ϵ ) ) log N , ( 3.9 ) h > \left( v - \dfrac{2k}{k+2\ell+1}\dfrac{2\ell+1}{\ell+1} \left( \dfrac{\vartheta}{2} - \epsilon \right) \right) \log N, \quad (3.9)
h > ( v − k + 2 ℓ + 1 2 k ℓ + 1 2 ℓ + 1 ( 2 ϑ − ϵ ) ) log N , ( 3.9 )
于是令ℓ = [ k / 2 ] \ell = [\sqrt{k} / 2] ℓ = [ k /2 ] k k k
h > ( v − 2 ϑ + 4 ϵ + O ( 1 k ) ) log N . ( 3.10 ) h > \left( v - 2\vartheta + 4\epsilon + O\left( \dfrac{1}{\sqrt{k}} \right) \right) \log N. \quad (3.10)
h > ( v − 2 ϑ + 4 ϵ + O ( k 1 ) ) log N . ( 3.10 )
于是得到了:
Δ r ≤ max ( v − 2 ϑ , 0 ) . ( 1.11 ) \Delta_r \le \max(v - 2\vartheta, 0). \quad (1.11)
Δ r ≤ max ( v − 2 ϑ , 0 ) . ( 1.11 )
而定理2即是r = 1 , ϑ = 1 / 2 r = 1, \vartheta = 1/2 r = 1 , ϑ = 1/2 证明完了定理2 .
定理1后半部分的证明
最后我们将证明定理1的后半部分.我们仍然是从S \mathcal{S} S L L L ℓ \ell ℓ L = o ( k ) L = o(k) L = o ( k )
S ′ : = ∑ n = N + 1 2 N ( ∑ i = 1 k θ ( n + h i ) − log 3 N ) ( ∑ ℓ = 0 L a ℓ Λ R ( n ; H k , ℓ ) ) 2 = ∑ n = N + 1 2 N ( ∑ i = 1 k θ ( n + h i ) − log 3 N ) × ∑ 0 ≤ ℓ 1 , ℓ 2 ≤ L a ℓ 1 a ℓ 2 Λ R ( n ; H k , ℓ 1 ) Λ R ( n ; H k , ℓ 2 ) = ∑ 0 ≤ ℓ 1 , ℓ 2 ≤ L a ℓ 1 a ℓ 2 M ℓ 1 ℓ 2 , ( 3.11 ) \begin{array}{ll}
\mathcal{S}' & := \displaystyle\sum_{n=N+1}^{2N} \left( \displaystyle\sum_{i=1}^{k} \theta(n+h_i) - \log 3N \right) \left( \displaystyle\sum_{\ell=0}^{L} a_\ell \Lambda_R(n; \mathcal{H}_k, \ell) \right)^2 \\
\\
& = \displaystyle\sum_{n=N+1}^{2N} \left( \displaystyle\sum_{i=1}^{k} \theta(n+h_i) - \log 3N \right) \\
\\
& \quad \times \displaystyle\sum_{0 \le \ell_1, \ell_2 \le L} a_{\ell_1}a_{\ell_2} \Lambda_R(n; \mathcal{H}_k, \ell_1) \Lambda_R(n; \mathcal{H}_k, \ell_2) \\
\\
& = \displaystyle\sum_{0 \le \ell_1, \ell_2 \le L} a_{\ell_1}a_{\ell_2} \mathcal{M}_{\ell_1\ell_2},
\end{array} \quad (3.11) S ′ := n = N + 1 ∑ 2 N ( i = 1 ∑ k θ ( n + h i ) − log 3 N ) ( ℓ = 0 ∑ L a ℓ Λ R ( n ; H k , ℓ ) ) 2 = n = N + 1 ∑ 2 N ( i = 1 ∑ k θ ( n + h i ) − log 3 N ) × 0 ≤ ℓ 1 , ℓ 2 ≤ L ∑ a ℓ 1 a ℓ 2 Λ R ( n ; H k , ℓ 1 ) Λ R ( n ; H k , ℓ 2 ) = 0 ≤ ℓ 1 , ℓ 2 ≤ L ∑ a ℓ 1 a ℓ 2 M ℓ 1 ℓ 2 , ( 3.11 )
其中我们有:
M ℓ 1 ℓ 2 = M ~ ℓ 1 ℓ 2 − ( log 3 N ) M ℓ 1 ℓ 2 , ( 3.12 ) \mathcal{M}_{\ell_1 \ell_2} = \widetilde{M}_{\ell_1 \ell_2} - (\log 3N) M_{\ell_1 \ell_2}, \quad (3.12)
M ℓ 1 ℓ 2 = M ℓ 1 ℓ 2 − ( log 3 N ) M ℓ 1 ℓ 2 , ( 3.12 )
注意到R = N ϑ 2 − ϵ R = N^{\frac{\vartheta}{2} - \epsilon} R = N 2 ϑ − ϵ M ℓ 1 ℓ 2 M_{\ell_1 \ell_2} M ℓ 1 ℓ 2 M ~ ℓ 1 ℓ 2 \widetilde{M}_{\ell_1 \ell_2} M ℓ 1 ℓ 2
M ℓ 1 ℓ 2 = ( ℓ 1 + ℓ 2 ℓ 1 ) ( log R ) k + ℓ 1 + ℓ 2 ( k + ℓ 1 + ℓ 2 ) ! S ( H k ) N , M_{\ell_1 \ell_2} = \dbinom{\ell_1 + \ell_2}{\ell_1} \dfrac{(\log R)^{k+\ell_1+\ell_2}}{(k+\ell_1+\ell_2)!} \mathfrak{S}(\mathcal{H}_k) N,
M ℓ 1 ℓ 2 = ( ℓ 1 ℓ 1 + ℓ 2 ) ( k + ℓ 1 + ℓ 2 )! ( log R ) k + ℓ 1 + ℓ 2 S ( H k ) N ,
M ~ ℓ 1 ℓ 2 = k ( ℓ 1 + ℓ 2 + 2 ℓ 1 + 1 ) ( log R ) k + ℓ 1 + ℓ 2 + 1 ( k + ℓ 1 + ℓ 2 + 1 ) ! S ( H k ) N . \widetilde M_{\ell_1 \ell_2} = k \dbinom{\ell_1+\ell_2+2}{\ell_1+1} \dfrac{(\log R)^{k+\ell_1+\ell_2+1}}{(k+\ell_1+\ell_2+1)!} \mathfrak{S}(\mathcal{H}_k) N.
M ℓ 1 ℓ 2 = k ( ℓ 1 + 1 ℓ 1 + ℓ 2 + 2 ) ( k + ℓ 1 + ℓ 2 + 1 )! ( log R ) k + ℓ 1 + ℓ 2 + 1 S ( H k ) N .
因此我们得到:
M ℓ 1 ℓ 2 ∼ ( ℓ 1 + ℓ 2 ℓ 1 ) S ( H k ) N ( log R ) k + ℓ 1 + ℓ 2 ( k + ℓ 1 + ℓ 2 ) ! × ( k ( ℓ 1 + ℓ 2 + 2 ) ( ℓ 1 + ℓ 2 + 1 ) ( ℓ 1 + 1 ) ( ℓ 2 + 1 ) ( k + ℓ 1 + ℓ 2 + 1 ) log R − log 3 N ) . \begin{array}{rl}
\mathcal{M}_{\ell_1 \ell_2} \sim & \dbinom{\ell_1+\ell_2}{\ell_1} \mathfrak{S}(\mathcal{H}_k) N \dfrac{(\log R)^{k+\ell_1+\ell_2}}{(k+\ell_1+\ell_2)!} \\
\\
& \times \left( \dfrac{k(\ell_1+\ell_2+2)(\ell_1+\ell_2+1)}{(\ell_1+1)(\ell_2+1)(k+\ell_1+\ell_2+1)} \log R - \log 3N \right).
\end{array} M ℓ 1 ℓ 2 ∼ ( ℓ 1 ℓ 1 + ℓ 2 ) S ( H k ) N ( k + ℓ 1 + ℓ 2 )! ( log R ) k + ℓ 1 + ℓ 2 × ( ( ℓ 1 + 1 ) ( ℓ 2 + 1 ) ( k + ℓ 1 + ℓ 2 + 1 ) k ( ℓ 1 + ℓ 2 + 2 ) ( ℓ 1 + ℓ 2 + 1 ) log R − log 3 N ) .
如果定义b ℓ = ( log R ) ℓ a ℓ b_\ell = (\log R)^\ell a_\ell b ℓ = ( log R ) ℓ a ℓ b \mathbf{b} b ( b 0 , b 1 , ⋯ , b L ) T (b_0, b_1, \cdots, b_L)^T ( b 0 , b 1 , ⋯ , b L ) T
S ∗ ( N , H k , ϑ , b ) : = 1 S ( H k ) N ( log R ) k + 1 S ′ ∼ ∑ 0 ≤ ℓ 1 , ℓ 2 ≤ L b ℓ 1 b ℓ 2 ( ℓ 1 + ℓ 2 ℓ 1 ) 1 ( k + ℓ 1 + ℓ 2 ) ! × ( k ( ℓ 1 + ℓ 2 + 2 ) ( ℓ 1 + ℓ 2 + 1 ) ( ℓ 1 + 1 ) ( ℓ 2 + 1 ) ( k + ℓ 1 + ℓ 2 + 1 ) − 2 ϑ ) ∼ b T M b , ( 3.13 ) \begin{array}{ll}
S^*(N, \mathcal{H}_k, \vartheta, \mathbf{b}) & := \dfrac{1}{\mathfrak{S}(\mathcal{H}_k) N (\log R)^{k+1}} S' \\
\\
& \sim \sum_{0 \le \ell_1, \ell_2 \le L} b_{\ell_1} b_{\ell_2} \dbinom{\ell_1+\ell_2}{\ell_1} \dfrac{1}{(k+\ell_1+\ell_2)!} \\
\\
& \quad \times \left( \dfrac{k(\ell_1+\ell_2+2)(\ell_1+\ell_2+1)}{(\ell_1+1)(\ell_2+1)(k+\ell_1+\ell_2+1)} - \dfrac{2}{\vartheta} \right) \\
\\
& \sim \mathbf{b}^T \mathbf{Mb},
\end{array} \quad (3.13) S ∗ ( N , H k , ϑ , b ) := S ( H k ) N ( log R ) k + 1 1 S ′ ∼ ∑ 0 ≤ ℓ 1 , ℓ 2 ≤ L b ℓ 1 b ℓ 2 ( ℓ 1 ℓ 1 + ℓ 2 ) ( k + ℓ 1 + ℓ 2 )! 1 × ( ( ℓ 1 + 1 ) ( ℓ 2 + 1 ) ( k + ℓ 1 + ℓ 2 + 1 ) k ( ℓ 1 + ℓ 2 + 2 ) ( ℓ 1 + ℓ 2 + 1 ) − ϑ 2 ) ∼ b T Mb , ( 3.13 )
其中
M = [ ( i + j i ) 1 ( k + i + j ) ! ( k ( i + j + 2 ) ( i + j + 1 ) ( i + 1 ) ( j + 1 ) ( k + i + j + 1 ) − 2 ϑ ) ] 0 ≤ i , j ≤ L . ( 3.14 ) \mathbf{M} = \left[ \dbinom{i+j}{i} \dfrac{1}{(k+i+j)!} \left( \dfrac{k(i+j+2)(i+j+1)}{(i+1)(j+1)(k+i+j+1)} - \dfrac{2}{\vartheta} \right) \right]_{0 \le i,j \le L}. \quad (3.14)
M = [ ( i i + j ) ( k + i + j )! 1 ( ( i + 1 ) ( j + 1 ) ( k + i + j + 1 ) k ( i + j + 2 ) ( i + j + 1 ) − ϑ 2 ) ] 0 ≤ i , j ≤ L . ( 3.14 )
而我们的目的是,选择一个b \mathbf{b} b ϑ \vartheta ϑ k k k S ∗ > 0 S^* > 0 S ∗ > 0 b \mathbf{b} b M \mathbf{M} M λ \lambda λ
S ∗ ∼ b T λ b = λ ∑ i = 0 k ∣ b i ∣ 2 . ( 3.15 ) S^* \sim \mathbf{b}^T \lambda \mathbf{b} = \lambda \sum_{i=0}^k |b_i|^2. \quad (3.15)
S ∗ ∼ b T λ b = λ i = 0 ∑ k ∣ b i ∣ 2 . ( 3.15 )
因此我们需要让M \mathbf{M} M 0 0 0
ϑ \vartheta ϑ k k k ℓ \ell ℓ h ( k ) h(k) h ( k )
1 1 1 6 6 6 1 1 1 16 16 16
0.95 0.95 0.95 7 7 7 1 1 1 20 20 20
0.90 0.90 0.90 8 8 8 2 2 2 26 26 26
0.85 0.85 0.85 10 10 10 2 2 2 32 32 32
0.80 0.80 0.80 12 12 12 2 2 2 42 42 42
0.75 0.75 0.75 16 16 16 2 2 2 60 60 60
0.70 0.70 0.70 22 22 22 4 4 4 90 90 90
0.65 0.65 0.65 35 35 35 4 4 4 158 158 158
0.60 0.60 0.60 65 65 65 6 6 6 336 336 336
0.55 0.55 0.55 193 193 193 9 9 9 1204 1204 1204
具体而言,当k = 6 , L = 1 , b 0 = 1 , b 1 = b k = 6,L = 1, b_0 = 1, b_1 = b k = 6 , L = 1 , b 0 = 1 , b 1 = b
S ∗ ∼ − 4 ( 1 − ϑ ) 8 ! ϑ ( ( b − 18 ϑ − 16 4 ( 1 − ϑ ) ) 2 + 15 ϑ 2 − 64 ϑ + 48 4 ( 1 − ϑ ) 2 ) . S^* \sim -\dfrac{4(1-\vartheta)}{8!\vartheta} \left(\left( b - \dfrac{18\vartheta - 16}{4(1 - \vartheta)} \right)^2 + \dfrac{15\vartheta^2 - 64\vartheta + 48}{4(1 - \vartheta)^2}\right).
S ∗ ∼ − 8 ! ϑ 4 ( 1 − ϑ ) ( ( b − 4 ( 1 − ϑ ) 18 ϑ − 16 ) 2 + 4 ( 1 − ϑ ) 2 15 ϑ 2 − 64 ϑ + 48 ) .
令b = 18 ϑ − 16 4 ( 1 − ϑ ) b = \frac{18\vartheta - 16}{4(1 - \vartheta)} b = 4 ( 1 − ϑ ) 18 ϑ − 16
S ∗ ∼ − 15 ϑ 2 − 64 ϑ + 48 8 ! ϑ ( 1 − ϑ ) , S^* \sim -\dfrac{15\vartheta^2 - 64\vartheta + 48}{8! \vartheta (1 - \vartheta)},
S ∗ ∼ − 8 ! ϑ ( 1 − ϑ ) 15 ϑ 2 − 64 ϑ + 48 ,
而要让S ∗ > 0 S^* > 0 S ∗ > 0
ϑ > 4 ( 8 − 19 ) 15 = 0.97096... ( 3.16 ) \vartheta > \dfrac{4(8 - \sqrt{19})}{15} = 0.97096... \quad (3.16)
ϑ > 15 4 ( 8 − 19 ) = 0.97096... ( 3.16 )
因此我们便成功证得了定理1的后半部分 .
总结
最后命题1和命题2的证明就直接承认了,还是有点太复杂了 .后面紧接着更新一集陈氏定理吧,趁此机会也再深入了解一遍Selberg筛法.以及本篇到底哪一块是和筛法联系起来的,我也得再学学Halberstam才行,我猜可能就是这个权函数,可能充当了特征的作用?唉唉唉,还得学!
(2025.4.18)
\quad
\quad 论文阅读 – Small gaps between primes (Maynard) 中稍微有所介绍),而这种筛法是对于admissible tuples进行筛选的,因此与Halberstam的那种对于大集合A \mathcal{A} A
\quad ω n \omega_n ω n ρ 1 \rho_1 ρ 1 ρ 2 \rho_2 ρ 2 筛法读书笔记(哥德巴赫猜想 by 潘承洞) – 加权筛法顶峰之陈景润定理 也有介绍)之间有没有联系呢?所代表的内涵是不是一致的呢?能不能结合在一块呢?而这些问题我暂时也还没有去研究,如果之后还有机会研究筛法的话再思考思考,但是很有可能毕设之后就是终章了.
