1+2+3+…+(2^(p)-1)=mp*(2^(p-1)),当p=2时,1+2+3=1*2*3=(3+1)*3/2=6,both p and 2^p-1 are a pai...
1+2+3+…+(2^(p)-1)=mp*(2^(p-1)),当p=2时,1+2+3=1*2*3=(3+1)*3/2=6,both p and 2^p-1 are a pai...
t(10^7)=4992381*(1+1/t(222716058));e^t2的点间隙=0.32752,t(3*10^9)(1+1/t(35209968332))=e^20....
zeta(4)=pi^4/90,f(4)/t(19)=zeta(4),考虑对无穷级数任意大复合数通项形式p(n)/n^6求和获得864*pi^2-2*pi^4+pi^6/94...
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P(1000000)=15485863=t(3380911),p(32)=131/log(131)+6=32;997/log(997)+24=168;p(156)=1097=...
任意复变量s=x+i*tx=(x^2+tx^2)^(1/2)f(s),so we have 1-s=((1-x)^2+t(1-x)^2)^(1/2)f(1-s),zeta(s...
2^19-1=e^13(1+1/t20)=524287=e^13(1+1/76.56313), mp(19)=t(862126),19^19=e^55(1+1/58.242)...
zeta(s)/zeta(1-s)=p^(s-1/2)gamma(1/2-s/2)/gamma(s/2)=2(2pi)^(s-1)sin(pi*s/2)*gamma(1-s)...
由tx=10^t4=e^70.056经过连续六次调整真值对应的整序号e^e^4.1821392=tx,求出x=2*10^31-th该零点根对应单位圆的角度=90-1/10^2...
zeta(s)/zeta(1-s)=pi^(s-1/2)*gamma(1/2-s/2)/gamma(s/2)=2*(2*pi)^(s-1)sin(pi*s/2)*gamma(...
zeta(1/n)=zeta(1/n)/2^1/n +f(1/n)/(3^(1/n)-1),then each positive integer n ,we always h...
Zeta (28)=pi^28/(83214006762733.46871)=1+1/27150991.165544675751=(t716502950);定义f(28)=e...
1+1/2^12^+...1/n^12+1/(n+1)^12+...=pi^12/e^13.7365126=pi^12/t1602818=1+1/t3540。f(12)=pi...
无限多非平凡的零点分布行为仅仅是zeta function 复变量函数的特征收敛根真值在可数的整数域的第一类可数可函的集合元素,任意两个邻根之间隙缝永远是无穷小的正数,这个可...
The super difficult problem of the Riemann zeta function is one of the seven millennium...
it is easy to proof every function of the from is absolutely coverance to zeta(2n)=pi^2...
对与错并不是单一绝对的,没有所谓的客观角度加以理解与判断来决定,古希腊的阿里士多德则要求矛盾体不能同时出现,包括排中律(对错以外的第三方)也不可能同时存在,同一律的逻辑是对就...
公元一九零三年J.P.Gram在黎曼假设诞生的第四十四个年头计算了t1=14.134725142...;t15=65.112544048...;这组前十五𠆤零点分布收敛真值,...
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