素数の規則を見つけたい。。。

素数の規則を見つけたい。。。 (701ﾚｽ)
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298: １３２人目の素数さん [sage] 2023/12/29(金) 21:13:50.10 ID:voXPt7J2

cos(2pi*(1/2+2/3+3/5))=cos(2pi*(7/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(11/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(13/(2*3*5)))=cos(2pi*((2*3*5-13)/(2*3*5)))=cos(2pi*(17/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(17/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(19/(2*3*5)))=cos(2pi*((2*3*5*7-19)/(2*3*5)))=cos(2pi*(41/(2*3*5)))
cos(2pi*(1/2+2/3+3/5))=cos(2pi*(23/(2*3*5)))
cos(2pi*(1/2+2/3+4/5))=cos(2pi*(29/(2*3*5)))
cos(2pi*(1/2+2/3+4/5))=cos(2pi*(31/(2*3*5)))
cos(2pi*(1/2+2/3+3/5))=cos(2pi*(37/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(41/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(43/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(47/(2*3*5)))

cos(2pi*(1/2+2/3+3/5+2/7))=cos(2pi*(11/(2*3*5*7)))=cos(2pi*((2*3*5*7-11)/(2*3*5*7)))=cos(2pi*(199/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+4/7))=cos(2pi*(13/(2*3*5*7)))=cos(2pi*((2*3*5*7-13)/(2*3*5*7)))=cos(2pi*(197/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+5/7))=cos(2pi*(17/(2*3*5*7)))=cos(2pi*((2*3*5*7-17)/(2*3*5*7)))=cos(2pi*(193/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+3/5+1/7))=cos(2pi*(19/(2*3*5*7)))=cos(2pi*((2*3*5*7-19)/(2*3*5*7)))=cos(2pi*(191/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+4/5+1/7))=cos(2pi*(23/(2*3*5*7)))=cos(2pi*((2*3*5*7-23)/(2*3*5*7)))=cos(2pi*(187/(2*3*5*7))) ←11*17
cos(2pi*(1/2+2/3+2/5+4/7))=cos(2pi*(29/(2*3*5*7)))=cos(2pi*((2*3*5*7-29)/(2*3*5*7)))=cos(2pi*(181/(2*3*5*7))) ←11^2以上、11^*17未満なので素数

http://rio2016.5ch.net/test/read.cgi/math/1640355175/298

315: １３２人目の素数さん [sage] 2023/12/31(日) 21:13:13.10 ID:ZQRjm/0R

(1/p(n)^(x+i*y')-1/p(n)^(x+i*y))=(2*1/p(n)^(x)*sin((y'-y)*lnp(n)/2)*e^(i*(arctan((-sin(y'*logp(n))+sin(ylogp(n)))/(cos(y'logp(n))-cos(ylogp(n))))+π)))

(1/2^(x+i*y')-1/2^(x+i*y))=(2*1/2^(x)*sin((y'-y)*ln2/2)*e^(i*(arctan((-sin(y'*log2)+sin(ylog2))/(cos(y'log2)-cos(ylog2)))+π)))
(1/3^(x+i*y')-1/3^(x+i*y))=(2*1/3^(x)*sin((y'-y)*ln3/2)*e^(i*(arctan((-sin(y'*log3)+sin(ylog3))/(cos(y'log3)-cos(ylog3)))+π)))
(1/4^(x+i*y')-1/4^(x+i*y))=(2*1/4^(x)*sin((y'-y)*ln4/2)*e^(i*(arctan((-sin(y'*log4)+sin(ylog4))/(cos(y'log4)-cos(ylog4)))+π)))

5^(1-x-i*y'))/(x-1+i*y')-5^(1-x-i*y)/(x-1+i*y)=5^(1-x)/√((x-1)^2+y'^2)*e^(i*'y'*ln5-arctan(y'/(x-1)))-5^(1-x)/√((x-1)^2+y^2)*e^(i*y*ln5-arctan(y/(x-1)))
5^(-(x+i*y'))/2-5^(-(x+i*y))/2=5^(-x)/2*(e^(i*-y'ln5)-e^(i*-yln5))

ζ(x+i*y')-ζ(x+i*y)≒(2*1/2^(x)*sin((y'-y)*ln2/2)*e^(i*(arctan((-sin(y'*log2)+sin(ylog2))/(cos(y'log2)-cos(ylog2)))+π)))
+(2*1/3^(x)*sin((y'-y)*ln3/2)*e^(i*(arctan((-sin(y'*log3)+sin(ylog3))/(cos(y'log3)-cos(ylog3)))+π)))
+(2*1/4^(x)*sin((y'-y)*ln4/2)*e^(i*(arctan((-sin(y'*log4)+sin(ylog4))/(cos(y'log4)-cos(ylog4)))+π)))
+5^(1-x)/√((x-1)^2+y'^2)*e^(i*'y'*ln5-arctan(y'/(x-1)))-5^(1-x)/√((x-1)^2+y^2)*e^(i*y*ln5-arctan(y/(x-1)))
+5^(-x)/2*(e^(i*-y'ln5)-e^(i*-yln5))
がx≠1/2のときy,y'をもたない(y≠y'>0)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/315

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