鲁思顺定理(一)

lusishun

定理:大于1717的偶数的哥猜素数对不少于20.
证明:由偶数的哥猜素数对公式:
4/9·4/2·6/4·8/6·9/7·……·40/38=20.361544，
去尾取整得20.

注:公式是由加强倍数含量两筛法得到的(省落).

wangyangke

http://www.mathchina.com/bbs/for ... &extra=page%3D4

lusishun

定理(二):
大于2209的偶数的哥猜素数对不少于24对。

cuikun-186

r2(1718)≥0.8488*1718/(ln1718)^2≈26.28


π(1718)=267，C(1718)=366


r2(1718)=366+2*267-1718/2=41


r2(1718)=41

cuikun-186

r2(2210)≥0.8488*2210/(ln2210)^2≈31.63

cuikun-186

验算一下10^5

众所周知，r2(10^5)=1620

r2(10^5)≥0.8488*10^5/(ln10^5)^2≈640.37

r2(10^5)=1620≥640逻辑严谨！


C(10^5)≥10^2/2+0.8488*10^5/(ln10^5)^2-2.51012*10^5/ln10^5≈28837.75

C(10^5)=32436

28837/32436≈0.889=88.9%


时空伴随者这只老妪母猴基因突变！看出笔误了！

cuikun-186

r2(1718)≥0.8488*1718/(ln1718)^2≈26.28


π(1718)=267，C(1718)=366


r2(1718)=366+2*267-1718/2=41


r2(1718)=41

单记分别是：

(19, 1699)
(61, 1657)
(97, 1621)
(109, 1609)
(139, 1579)
(151, 1567)
(229, 1489)
(271, 1447)
(337, 1381)
(397, 1321)
(421, 1297)
(439, 1279)
(487, 1231)
(547, 1171)
(601, 1117)
(631, 1087)
(709, 1009)
(727, 991)
(751, 967)
(811, 907)
(859, 859)

双记法分别是：
(19,1699) -> (19,1699), (1699,19)
(61,1657) -> (61,1657), (1657,61)
(97,1621) -> (97,1621), (1621,97)
(109,1609) -> (109,1609), (1609,109)
(139,1579) -> (139,1579), (1579,139)
(151,1567) -> (151,1567), (1567,151)
(229,1489) -> (229,1489), (1489,229)
(271,1447) -> (271,1447), (1447,271)
(337,1381) -> (337,1381), (1381,337)
(397,1321) -> (397,1321), (1321,397)
(421,1297) -> (421,1297), (1297,421)
(439,1279) -> (439,1279), (1279,439)
(487,1231) -> (487,1231), (1231,487)
(547,1171) -> (547,1171), (1171,547)
(601,1117) -> (601,1117), (1117,601)
(631,1087) -> (631,1087), (1087,631)
(709,1009) -> (709,1009), (1009,709)
(727,991) -> (727,991), (991,727)
(751,967) -> (751,967), (967,751)
(811,907) -> (811,907), (907,811)
(859,859) -> (859,859) 只一次

cuikun-186

r2(N)≥0.8488N/(lnN)^2


r2(2210)≥0.8488*2210/(ln2210)^2≈31.63

cuikun-186

π(2210)=329，C(2210)=541


r2(2210)=541+2*329-2210/2=94


r2(2210)=94

lusishun

大路朝天，各走一边。远离为好，免的争执，
大于1717的偶数的哥猜素数对不少于20对，看来没有瑕疵。谢谢