孿生質數者,質數之差二也,若三與五,五與七爾爾。 3, 5, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, etc 難題 孿生質數之無窮乎。遂有數學家究之,然求其倒數之和,值收斂,名曰布隆常數[一] 1 3 + 1 5 + 1 5 + 1 7 + 1 11 + 1 13 + ⋯ ⋯ = 1.90216058 ⋯ ⋯ {\displaystyle {\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{11}}+{\frac {1}{13}}+\cdots \cdots =1.90216058\cdots \cdots } 今仍無解,或下文爲可證其無窮性。 Remove ads無窮性之證 全部證明請見條目之孪生质数无穷多的证明者也. N(twin primes) = lim n → ∞ [ n ⋅ 1 3 ( 1 − 2 5 ) ( 1 − 2 7 ) ( 1 − 2 11 ) ( 1 − 2 13 ) . . . . . . ] {\displaystyle \mathrm {=\lim _{n\to \infty }[n\cdot {\frac {1}{3}}(1-{\frac {2}{5}})(1-{\frac {2}{7}})(1-{\frac {2}{11}})(1-{\frac {2}{13}})......]} } = lim n → ∞ [ n ⋅ ( 1 − 2 3 ) ( 1 − 2 5 ) ( 1 − 2 7 ) ( 1 − 2 11 ) ( 1 − 2 13 ) . . . . . . ] {\displaystyle \mathrm {=\lim _{n\to \infty }[n\cdot (1-{\frac {2}{3}})(1-{\frac {2}{5}})(1-{\frac {2}{7}})(1-{\frac {2}{11}})(1-{\frac {2}{13}})......]} } = lim n → ∞ n ⋅ k [ ( 1 − 1 3 ) ( 1 − 1 5 ) ( 1 − 1 7 ) ( 1 − 1 11 ) ( 1 − 1 13 ) . . . . . . ] 2 {\displaystyle \mathrm {=\lim _{n\to \infty }n\cdot k[(1-{\frac {1}{3}})(1-{\frac {1}{5}})(1-{\frac {1}{7}})(1-{\frac {1}{11}})(1-{\frac {1}{13}})......]^{2}} } = lim n → ∞ 4 k n [ ( 1 − 1 2 ) ( 1 − 1 3 ) ( 1 − 1 5 ) ( 1 − 1 7 ) ( 1 − 1 11 ) ( 1 − 1 13 ) . . . . . . ] 2 {\displaystyle \mathrm {=\lim _{n\to \infty }4kn[(1-{\frac {1}{2}})(1-{\frac {1}{3}})(1-{\frac {1}{5}})(1-{\frac {1}{7}})(1-{\frac {1}{11}})(1-{\frac {1}{13}})......]^{2}} } = lim n → ∞ 4 k n [ ln ( n ) + γ ] 2 {\displaystyle \mathrm {=\lim _{n\to \infty }{\frac {4kn}{[\ln(n)+\gamma ]^{2}}}} } = ∞ {\displaystyle \mathrm {=\infty } } Remove ads兼查 senary duodecimal 據Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads