太阳 [(10^5k+1)/(10^k+1)] / [(10^n+1)/11]问题

yangchuanju

太阳  [(10^5k+1)/(10^k+1)] / [(10^n+1)/11]问题

已经得到的(10^5k+1)/(10^k+1)分解式有：
(10^5+1)/(10^1+1) =  9091
(10^10+1)/(10^2+1) =  3541 × 27961  =  99009901
(10^15+1)/(10^3+1) =  211 × 241 × 2161 × 9091  =  999000999001
(10^20+1)/(10^4+1) =  1676321 × 5964848081<10>  =  9999000099990001
(10^25+1)/(10^5+1) =  251 × 5051 × 78875943472201<14>  =  99999000009999900001
(10^30+1)/(10^6+1) =  61 ×3541 × 27961 × 4188901 × 39526741  =  999999000000999999000001
(10^35+1)/(10^7+1) =  9091 × 4147571 × 265212793249617641<18>  =  9999999000000099999990000001
(10^40+1)/(10^8+1) =  5070721 × 19721061166646717498359681<26>  =  99999999000000009999999900000001
(10^45+1)/(10^9+1) =  211 × 241 × 2161 × 9091 × 29611 × 3762091 × 8985695684401<13>
(10^50+1)/(10^10+1) =  60101 × 7019801 × 14103673319201<14> × 1680588011350901<16>
(10^55+1)/(10^11+1) =  331 × 5171 × 9091 × 20163494891<11> × 318727841165674579776721<24>
(10^60+1)/(10^12+1) =  1676321 × 5964848081<10> × 100009999999899989999000000010001<33>
(10^65+1)/(10^13+1) =  131 × 9091 × 8396862596258693901610602298557167100076327481<46>
(10^70+1)/(10^14+1) =  421 × 3541 × 27961 × 3471301 × 13489841 × 60368344121<11> × 848654483879497562821<21>
(10^75+1)/(10^15+1) =  251 × 5051 × 78875943472201<14> × 10000099999999989999899999000000000100001<41>
(10^80+1)/(10^16+1) =  1634881 × 18453761 × 947147262401<12> × 349954396040122577928041596214187605761<39>
(10^85+1)/(10^17+1) =  9091 × 87211 × 787223761 × 1602207948210144520667419183035809176643&#172;86555934641<51>
(10^90+1)/(10^18+1) =  61 × 181 × 3541 × 27961 × 4188901 × 39526741 × 4999437541453012143121<22> × 1105097795002994798105101<25>
(10^95+1)/(10^19+1) =  9091 × 1812604116731<13> × 121450506296081<15> × 4996731930447843676185843959746621491531100801<46>
(10^100+1)/(10^20+1) =  401 × 1201 × 1601 × 129694419029057750551385771184564274499075700947656757821537291527196801<72>
它们的分解式中间有不少个都含有素因子9091，可被(10^5+1)/11 =9091整除，9091是素数。

除数(10^n+1)/11中的n只能是奇数，只取其中所有素因子尾数全是1的有：
(10^5+1)/11 =9091
(10^7+1)/11 =909091
(10^19+1)/11 =909090909090909091<18>
(10^25+1)/11 =251×5051×9091×78875943472201<14>=909090909090909090909091<c24>
(10^31+1)/11 =909090909090909090909090909091<30>
(10^53+1)/11 =9090909090909090909090909090909090909090909090909091<52>
(10^61+1)/11 =81131×11205222530116836855321528257890437575145023592596037161<56>
=909090909090909090909090909090909090909090909090909090909091<c60>
(10^67+1)/11 =909090909090909090909090909090909090909090909090909090909090909091<66>
(10^95+1)/11 =9091×1812604116731×121450506296081<15>×909090909090909091<18>×4996731930447843676185843959746621491531100801<46>
=9090909090909090909090909090909090909090909090909090909090909090909090909090909090909090909091<c94>
其中n=5,7,19,31,53,67时(10^n+1)/11各含一个90…91型素数因子（注意则几个指数n则都是素数）；当n=25,95时(10^n+1)/11不是素数，但其乘积是一个90…91型合数。

(10^5k+1)/(10^k+1)分解式中的各个素因子全是尾数等于1的素数，许多分解式都含有素因子9091、251、5051等；
(10^n+1)/11分解式中含有素因子9091、251、5051的只有(10^5+1)/11 =9091、(10^25+1)/11 =251×5051×9091×78875943472201<14>、(10^95+1)/11 =9091×1812604116731×121450506296081<15>×……等寥寥无几的几个。
经试算，(10^25+1)/(10^5+1) =251 × 5051× 78875943472201<14>，再除以(10^25+1)/11 =251×5051×9091×78875943472201<14>时中间多了一个因子9091，不能整除；
(10^125+1)/(10^25+1) =21001×162251×10893295001<11> ×2694097928…1<81>不能被(10^25+1)/11 =251×5051×9091×78875943472201<14>或(10^5+1)/11 =9091整除；
同样(10^95+1)/11也不能整除(10^5k+1)/(10^k+1)。

据此分析，在进行二次相除并且要能整除时，只能用(10^n+1)/11中的 (10^5+1)/11 =9091作除数， 9091是素数，
[(10^5k+1)/(10^k+1)] / [(10^n+1)/11] 中的 (10^n+1)/11“必定是素数”！

太阳先生梦寐以求的素数公式终于找到了，在此特向太阳先生祝贺！
也请太阳先生尽快整理资料，撰写论文，争取早日获得“诺贝尔数学奖”！

白新岭

如果方法不错的话，那真是个数学奇迹。

白新岭

我在波浪的QQ空间中看到过素数公式（不是求数量，而是给出素数），那是一个含有复数和三角函数形式的公式，切采用了指数形式。

太阳

yangchuanju：找到反例吗？没有找到反例吧！