素数の規則を見つけたい。。。

241: １３２人目の素数さん [sage] 2023/12/23(土) 02:17:03.83 ID:O5dB6rNY

>>240

(((a^2+b^2)*e^(i*2*arctan(a/b))+2^(3/2)*a*b*e^(i*-π/4)))=-a^2+2ab+b^2
(((a^2+b^2)*e^(i*2*arctan(b/a))+2^(3/2)*a*b*e^(i*-π/4)))=a^2+2ab-b^2

(n+1)^2+2n*(n+1)-n^2=(n+1)^2+2(n+2)*(n+1)-(n+2)^2 ←nに何を入れても等しくなる

(((2^2+1^2)*e^(i*2*arctan(2/1))+2^(3/2)*2*1*e^(i*-π/4)))=1
(((2^2+1^2)*e^(i*2*arctan(1/2))+2^(3/2)*2*1*e^(i*-π/4)))=7

(((2^2+3^2)*e^(i*2*arctan(3/2))+2^(3/2)*2*3*e^(i*-π/4)))=7
(((2^2+3^2)*e^(i*2*arctan(2/3))+2^(3/2)*2*3*e^(i*-π/4)))=17

(((4^2+3^2)*e^(i*2*arctan(4/3))+2^(3/2)*3*4*e^(i*-π/4)))=17
(((4^2+3^2)*e^(i*2*arctan(3/4))+2^(3/2)*3*4*e^(i*-π/4)))=31

(((4^2+5^2)*e^(i*2*arctan(5/4))+2^(3/2)*5*4*e^(i*-π/4)))=31
(((4^2+5^2)*e^(i*2*arctan(4/5))+2^(3/2)*5*4*e^(i*-π/4)))=49

(((6^2+5^2)*e^(i*2*arctan(6/5))+2^(3/2)*5*6*e^(i*-π/4)))=49
(((6^2+5^2)*e^(i*2*arctan(5/6))+2^(3/2)*5*6*e^(i*-π/4)))=71

ζ1(s)=|ζ1(s)|*e^(i*θ)      ←素数のみのゼータ関数(s=0点の時)=1/2^s+1/3^s+1/5^s+・・・
ζ2(s)=|ζ2(s)|*e^(i*(θ+π))　←非素数のみのゼータ関数(s=0点の時)=1+1/4^s+1/6^s+1/8^s+1/9^s+・・・

ζ1(s)+ζ2(s)=ζ(s)
|ζ1(s)|=|ζ2(s)|
ζ1(s)*ζ2(s)=|ζ1(s)|*|ζ2(s)|*e^(i*(2θ+π))

ζ1(s)=√(ζ1(s)*ζ2(s))*e^(-iπ/2)
ζ2(s)=√(ζ1(s)*ζ2(s))*e^(iπ/2)

√(ζ1(s)*ζ2(s))*e^(-iπ/2)+√(ζ1(s)*ζ2(s))*e^(iπ/2)=ζ(s)

(√(ζ1(s)*ζ2(s))*e^(-iπ/2)+√(ζ1(s)*ζ2(s))*e^(iπ/2))^2=ζ(s)^2

(ζ1(s)*ζ2(s))*e^(-iπ)+(ζ1(s)*ζ2(s))*e^(iπ)+2*(ζ1(s)*ζ2(s))=ζ(s)^2 ←2*(ζ1(s)*ζ2(s))=ζ(s)^2になる

2^n*(ζ1(s)*ζ2(s))^(1/n)=2*(ζ1(s)*ζ2(s))=(ζ1(s)+ζ2(s))^2=ζ(s)^2

lim [n→∞]　2^(n-1)*(ζ1(s)*ζ2(s))^(1/n)=ζ(s) ←n→無限のとき

http://rio2016.5ch.net/test/read.cgi/math/1640355175/241