孪生质数无穷多的证明 - johanzumimvon/Johan-zumimvon-Christianity GitHub Wiki

孪生质数, 亦名ユポㇷ゚リムㇲ, 兄弟质数, 是指相距为2的二个质数(ㇷ゚リムㇲ), 比如5与7, 11与13(十二进制之作#与11), 17与19(十二进制之作15与17), 29与31(十二进制之作25与27).

孪生质数列表(十进制)

1, 3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73, 101, 103, 107, 109, 137, 139, 149, 151, 179, 181, 191, 193, 197, 199, 227, 229, 239, 241, 269, 271, 281, 283, 311, 313, 347, 349, 419, 421, 431, 433, 461, 463, 521, 523, 569, 571, 599, 601, 617, 619, 641, 643

孪生质数列表十二进制版

1, 3, 5, 7, E, 11, 15, 17, 25, 27, 35, 37, 4E, 51, 5E, 61, 85, 87, 8E, 91, E5, E7, 105, 107, 12E, 131, 13E, 141, 145, 147, 16E, 171, 17E, 181, 1X5, 1X7, 1E5, 1E7, 21E, 221, 24E, 251, 2XE, 2E1, 2EE, 301, 325, 327, 375, 377, 3E5, 3E7, 41E, 421, 435, 437, 455, 457, 46E, 471, 575, 577, 585, 587, 58E, 591, 5E5, 5E7, 615, 617, 70E, 711, 71E, 721, 735, 737, 745, 747, 76E, 771, 7EE, 801, 865, 867, 8X5, 8X7, 8E5, 8E7, 905, 907, 91E, 921, 9XE, 9E1, X0E, X11, X35, X37, X3E, X41, E1E, E21, E2E, E31, E6E, E71, E95, E97, EE5, EE7

引理

微分、导函数

由微分可知

$\mathrm{\frac{dx^2}{dx}}$

$\mathrm{=\frac{(x+\Delta x)^2-x^2}{\Delta x}}$

$\mathrm{=\frac{2x\Delta x+(\Delta x)^2}{\Delta x}}$

$\mathrm{=2x+\Delta x}$

$\mathrm{\Delta x}$ 趋于0时， $\mathrm{\frac{dx^2}{dx}}$ 会趋于2x, 遂有极限式:

$\mathrm{\lim_{n \to \infty} (1-\frac{2}{n})=\lim_{n \to \infty} (1-\frac{1}{n})^2}$

尤拉乘积

由ユラ乘积可知

$\mathrm{\frac{1}{\lim_{n\to\infty}[\ln(n)+\gamma]}}$=[(1−2⁻¹)(1−3⁻¹)(1−5⁻¹)(1−7⁻¹)·etc]

来自尤拉乘积的推论

通俗的说, 就是指:

$\mathrm{(1-\frac{2}{n}) \approx (1-\frac{1}{n})^2}$ (n趋于无穷大时, 二者相等)

遂有

$\mathrm{\lim_{n \to \infty} \frac{n}{[\ln(n)]^2}=\infty}$

$\mathrm{(1-\frac{1}{n})^2 \approx (1-\frac{2}{n})}$

证明

N(twin primes)

$\mathrm{=\lim_{n \to \infty}[n \cdot \frac{1}{3} (1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})......]}$

$\mathrm{=\lim_{n \to \infty}[n \cdot (1-\frac{2}{3})(1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})......]}$

$\mathrm{=\lim_{n \to \infty}n \cdot k[(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})......]^2}$

$\mathrm{=\lim_{n \to \infty}4kn[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})......]^2}$

$\mathrm{=\lim_{n \to \infty} \frac{4kn}{[\ln(n)+\gamma]^2}}$

$=\infty$

也就是孪生质数有无穷多.

意味

孪生质数无穷多意味着有无穷多个形如(6n−1)的质数与形如(6n+1)的质数, 因为除了3, 所有的孪生质数相邻于6n.

推论

波利尼亚克(Polignac, ポリナ゙ㇰ)猜想也可被如此方法所证明.

ポリナ゙ㇰ猜想

对所有自然数k, 存在无穷多个素数对, 也就是

p与(p+2k)

比如k=2时为3, 7, 11, 13, 17, 19, 23, 37, 41, 43, 47, 67, 71, 79, 83, 97, 101, 103, 107, 109, 113, 127, 131, 163, 167, 193, 197, 223, 227, 229, 233, 277, 281, 307, 311, 313, 317, 349, 353, 379, 383, 397, 401, 439, 443, 457, 461, 463, 467, 487, 491, 499, 503, 613, 617, 643

十二进制则为

3, 7, E, 11, 15, 17, 1E, 31, 35, 37, 3E, 57, 5E, 67, 6E, 81, 85, 87, 8E, 91, 95, X7, XE, 117, 11E, 141, 145, 167, 16E, 171, 175, 1E1, 1E5, 217, 21E, 221, 225, 251, 255, 277, 27E, 291, 295, 307, 30E, 321, 325, 327, 32E, 347, 34E, 357, 35E, 431, 435, 457, 45E, 481, 485, 517, 51E, 531, 535, 541, 545, 587, 58E, 5E1, 5E5, 5E7, 5EE, 611, 615, 617, 61E, 637, 63E, 661, 665, 687, 68E, 701, 705, 767, 76E, 771, 775, 851, 855, 8X7, 8XE, 901, 905, 907, 90E, 9X7, 9XE, 9E1, 9E5, X07, X0E, X37, X3E, X41, X45, X91, X95, XX7, XXE, XE7, XEE, E11, E15, E21, E25, E67, E6E, E91, E95

也可以被相同的方法证明有无穷多个.

ポリナ゙ㇰ质数 | ポリナ゙ㇰㇷ゚リムㇲ

k=3

5, E, 7, 11, E, 15, 11, 17, 15, 1E, 1E, 25, 27, 31, 31, 37, 35, 3E, 3E, 45, 45, 4E, 51, 57, 57, 61, 61, 67, 6E, 75, 81, 87, 85, 8E, 87, 91, 8E, 95, XE, E5, 107, 111, 111, 117), 11E, 125, 125, 12E, 13E, 145, 141, 147, 167, 171, 16E, 175, 175, 17E, 18E, 195, 195, 19E, 19E, 1X5, 1X7, 1E1, 1E1, 1E7, 217, 221, 21E, 225, 237, 241, 24E, 255, 255, 25E, 267, 271, 271, 277, 27E, 285, 301, 307, 30E, 315, 321, 327, 325, 32E, 35E, 365, 391, 397, 3X5, 3XE, 3XE, 3E5, 3E7, 401, 40E, 415, 415, 41E, 421, 427, 427, 431, 431, 437, 455, 45E, 45E, 465, 465, 46E, 485, 48E, 507, 511, 511, 517, 527, 531, 585, 58E, 587, 591, 5E1, 5E7, 5E5, 5EE, 611, 617, 615, 61E, 665, 66E, 66E, 675, 68E, 695, 695, 69E, 6X7, 6E1, 705, 70E, 721, 727, 747, 751, 767, 771, 76E, 775, 775, 77E, 77E, 785, 791, 797, 797, 7X1, 825, 82E, 82E, 835, 855, 85E, 85E, 865, 867, 871, 8X5, 8XE, 8XE, 8E5, 8E7, 901, 901, 907, 905, 90E, 921, 927, 955, 95E, 95E, 965, 9X7, 9E1, 9XE, 9E5, 9E5, 9EE, X07, X11, X11, X17, X35, X3E, X37, X41, X3E, X45, X45, X4E, X87, X91, X95, X9E, E15, E1E, E1E, E25, E25, E2E, E31, E37, E61, E67, E67, E71, E91, E97

k=4

3, E, 5, 11, E, 17, 1E, 27, 25, 31, 45, 51, 4E, 57, 5E, 67, 75, 81, 85, 91, XE, E7, 105, 111, 125, 131, 13E, 147, 175, 181, 19E, 1X7, 1X5, 1E1, 25E, 267, 285, 291, 295, 2X1, 2EE, 307, 315, 321, 33E, 347, 34E, 357, 3XE, 3E7, 3E5, 401, 415, 421, 41E, 427, 465, 471, 48E, 497, 4X5, 4E1, 4EE, 507, 51E, 527, 535, 541, 585, 591, 63E, 647, 655, 661, 69E, 6X7, 705, 711, 71E, 727, 745, 751, 785, 791, 80E, 817, 835, 841, 85E, 867, 865, 871, 8XE, 8E7, 8E5, 901, 91E, 927, 965, 971, 9EE, X07, X0E, X17, X35, X41, X6E, X77, X9E, XX7, XXE, XE7, E15, E21, E25, E31, E2E, E37, 

小百科

n以内孪生质数的估计

$\mathrm{\pi_2(n)\approx \frac{4}{3}\frac{n}{[\ln(n)]^2}}$

质数六元组有无穷多个

质数六元组, 亦名prime sextuplet. 是指以下数:

7, 11, 13, 17, 19, 23, 97, 101, 103, 107, 109, 113, 16057, 16061, 16063, 16067, 16069, 16073, 19417, 19421, 19423, 19427, 19429, 19433, 43777, 43781, 43783, 43787, 43789, 43793, 1091257, 1091261, 1091263, 1091267, 1091269, 1091273, 1615837, 1615841, 1615843, 1615847, 1615849, 1615853,

十二进制为

7, #, 11, 15, 17, 1#, 81, 85, 87, 8#, 91, 95, 9361, 9365, 9367, 936#, 9371, 9375, #2∗1, #2∗5, #2∗7, #2∗#, #2#1, #2#5, 21401, 21405, 21407, 2140#, 21411, 21415, 447621, 447625, 447627, 44762#, 447631, 447635, 65#111, 65#115, 65#117, 65#11#, 65#121, 65#125, etcetera(エㇳケテラ)

六元质数组有初始质数则有

7, 81, 9361, #2∗1, 21401, 447621, 65#111, 7∗2##1, #41617, #4#587, 1104#21, 117#867, 2017767, 2217797, 24∗7707, 2686107, 2849611, 2#16641, 3637911, 3867##1, 3∗∗6397, 4∗43117

由于质数六元组受制于2, 3, 5, 7这四个质数, 也就是除了7, 11, 13, 17, 19, 23这六个初始者, 其他的质数六元组者可以表示成210n+97, 210n+101, 210n+103, 210n+107, 210n+109, 210n+113.

这样, 质数六元组的数目应该为

$\mathrm{\frac{6n}{210}(1−\frac{6}{11})(1−\frac{6}{13})(1−\frac{6}{17})(1−\frac{6}{19})(1−\frac{6}{23})(1−\frac{6}{29})\cdot etc}$

$\mathrm{=\frac{n}{35}(1−\frac{6}{11})(1−\frac{6}{13})(1−\frac{6}{17})(1−\frac{6}{19})(1−\frac{6}{23})(1−\frac{6}{29})\cdot etc}$

$\mathrm{=kn[(1−\frac{1}{2})(1−\frac{1}{3})(1−\frac{1}{5})(1−\frac{1}{7})(1−\frac{1}{11})(1−\frac{1}{13})(1−\frac{1}{17})(1−\frac{1}{19})(1−\frac{1}{23})(1−\frac{1}{29})\cdot etc]^{6}}$

$\mathrm{=\lim_{n\to\infty}\cdot\frac{kn}{[\ln(n)]^{6}}}$

$=\infty$

也就是质数六元组有无穷多个

推论

质数四元组有无穷多个.

质数四元组亦名prime quadruplets, 有如下二种形式

5, 7, E, 11, E, 11, 15, 17, 85, 87, 8E, 91, 13E, 141, 145, 147, 585, 587, 58E, 591, X35, X37, X3E, X41, 10EE, 1101, 1105, 1107, 1255, 1257, 125E, 1261, 1X6E, 1X71, 1X75, 1X77, 2005, 2007, 200E, 2011, 332E, 3331, 3335, 3337, 555E, 5561, 5565, 5567, 7635, 7637, 763E, 7641, 9075, 9077, 907E, 9081, 912E, 9131, 9135, 9137, 9365, 9367, 936E, 9371, X535, X537, X53E, X541, XE3E, XE41, XE45, XE47, E2X5, E2X7, E2XE, E2E1

7, E, 11, 15, 11, 15, 17, 1E, 31, 35, 37, 3E, 81, 85, 87, 8E, 87, 8E, 91, 95, 167, 16E, 171, 175, 217, 21E, 221, 225, 321, 325, 327, 32E, 5E1, 5E5, 5E7, 5EE, 611, 615, 617, 61E, 767, 76E, 771, 775, 901, 905, 907, 90E, 9X7, 9XE, 9E1, 9E5, X37, X3E, X41, X45, 10E7, 10EE, 1101, 1105, 11X1, 11X5, 11X7, 11XE, 1677, 167E, 1681, 1685, 2001, 2005, 2007, 200E, 2741, 2745, 2747, 274E, 2927, 292E, 2931, 2935, 3037, 303E, 3041, 3045, 3327, 332E, 3331, 3335, 3E47, 3E4E, 3E51, 3E55, 4681, 4685, 4687, 468E, 4967, 496E, 4971, 4975, 6071, 6075, 6077, 607E, 7E07, 7E0E, 7E11, 7E15, 8041, 8045, 8047, 804E, 9127, 912E, 9131, 9135, 9361, 9365, 9367, 936E, 9367, 936E, 9371, 9375, 9447, 944E, 9451, 9455, X087, X08E, X091, X095, E2X1, E2X5, E2X7, E2XE, E2X7, E2XE, E2E1, E2E5

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