黎曼zeta函数与分圆整数内在深刻联系
ζ ( s ) = ∑ x = 1 ∞ 1 x s = ∏ m = 0 n − 1 1 1 − e i 2 m π n ( ( 2 − s ) + ∑ k = 1 ∞ p k − s ∏ j = 0 k − 1 ( 1 − p j − s ) ) 1 n {\displaystyle \zeta (s)=\sum _{x=1}^{\infty }{\frac {1}{x^{s}}}=\prod _{m=0}^{n-1}{\cfrac {1}{1-{e^{i{\tfrac {2m\pi }{n}}}}{{\Big (}(2^{-s})+{\displaystyle \sum _{k=1}^{\infty }}p_{k}^{-s}{\displaystyle \prod _{j=0}^{k-1}}{(1-p_{j}^{-s})}}{\Big )}^{\tfrac {1}{n}}}}\,}
ζ ( s ) = ∑ x = 1 ∞ 1 x s = 1 n ∑ m = 0 n − 1 1 1 − e i 2 m π n ( ( 2 − s ) + ∑ k = 1 ∞ p k − s ∏ j = 0 k − 1 ( 1 − p j − s ) ) 1 n {\displaystyle \zeta (s)=\sum _{x=1}^{\infty }{\frac {1}{x^{s}}}={\frac {1}{n}}\sum _{m=0}^{n-1}{\cfrac {1}{1-{e^{i{\tfrac {2m\pi }{n}}}}{{\Big (}(2^{-s})+{\displaystyle \sum _{k=1}^{\infty }}p_{k}^{-s}{\displaystyle \prod _{j=0}^{k-1}}{(1-p_{j}^{-s})}}{\Big )}^{\tfrac {1}{n}}}}}
p 0 = 2 , p 1 = 3 , p 2 = 5 , p 3 = 7 , . . . , p i , . . . {\displaystyle p_{0}=2,p_{1}=3,p_{2}=5,p_{3}=7,...,p_{i},...}
p i ∈ p r i m e s {\displaystyle p_{i}\in primes} in order
∀ n ≥ 1 , ∈ Z {\displaystyle \forall n\geq 1,\in \mathbb {Z} }
显然该公式当n=1和n=2时简化为著名的欧拉连乘积形式，但n=3时的简化形式呢,n=4时或等于后续整数时的简化形式呢，这引起过您深入之思考吗？
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