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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181: １３２人目の素数さん [sage] 2023/09/17(日) 00:45:26.83 ID:NvL18fxN

e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (n/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))
cos(2  π (2/17(2/13(2/11 (2/7 (2/5 (n/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2)) >cos(2π*19^2/(210*11*13*17))
1/32 (255255 m - 145721)<n<5/32 (51051 m - 29072)

e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (3433/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((283 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (-4546/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((137 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (3430/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((91 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (-4547/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((73 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (3428/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((-37 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (3427/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((-101 i π)/255255)
e^(2 i π (2/17(2/13(2/11 (2/7 (2/5 (3425/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)+1/2))=e^((-229 i π)/255255)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/181

201: １３２人目の素数さん [sage] 2023/11/26(日) 00:35:25.83 ID:5ylX1SN5

↓３次元では書けないベクトル和（((x+y+i^m*z)*(x+y-i^m*z)*(x-y+i^m*z)*(x-y-i^m*z)) mが３以上のベクトル和をかけない）　
√(x^4 - 2 x^2 y^2 + 2 x^2 z^2 + y^4 + 2 y^2 z^2 + z^4)=√(((x+y+i^(2n+1)*z)*(x+y-i^(2n+1)*z)*(x-y+i^(2n+1)*z)*(x-y-i^(2n+1)*z)))　　　
√(x^4 - 2 x^2 y^2 - 2 x^2 z^2 + y^4 - 2 y^2 z^2 + z^4)=√(((x+y+i^2n*z)*(x+y-i^2n*z)*(x-y+i^2n*z)*(x-y-i^2n*z)))

cos(2pi*((2*a+1)/2^3-(3*b+1)/3^3-c/5^3-d/7^3+e/11^3)) > cos(2pi*(13^2/2310^3))

a = 4 n_1, b = 9 n_2, c = 125 n_3, d = 343 n_4 + 83, e = 1331 n_5 + 205,　
a = 4 n_1, b = 9 n_2, c = 125 n_3 + 53, d = 7 (49 n_4 + 29), e = 1331 n_5 + 1235, cos(2pi*((2*4+1)/2^3-(3*9+1)/3^3-53/5^3-7*29/7^3+1235/11^3))=cos((91 π)/6163195500) ←7*29が7をもつため非素数
a = 4 n_1, b = 3 (3 n_2 + 1), c = 125 n_3 + 77, d = 343 n_4 + 163, e = 1331 n_5 + 448, 　cos(2pi*((2*4+1)/2^3-(3*3+1)/3^3-77/5^3-163/7^3+448/11^3))＝cos((19 π)/6163195500)
a = 4 n_1, b = 3 (3 n_2 + 2), c = 125 n_3 + 29, d = 343 n_4 + 243, e = 1331 n_5 + 691,　cos(2pi*((2*4+1)/2^3-(3*6+1)/3^3-29/5^3-243/7^3+691/11^3))=cos((163 π)/6163195500)
a = 4 n_1, b = 3 (3 n_2 + 2), c = 125 n_3 + 101, d = 343 n_4 + 123, e = 1331 n_5 + 992,  cos(2pi*((2*4+1)/2^3-(3*6+1)/3^3-101/5^3-123/7^3+992/11^3))=cos((53 π)/6163195500)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/201

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

312: １３２人目の素数さん [sage] 2023/12/31(日) 14:52:49.83 ID:ZQRjm/0R

ゼータ関数をζ(x+i*y)≒1+1/2^(1/2+i*y)+1/3^(1/2+i*y)+1/4^(1/2+i*y)と簡略化
ζ(x+i*y')=ζ(x+i*y)となるときゼロ点しかないと仮定する(y'≠y)

|半径1/P(n)^(x/2^m)の円内の余弦の長さ|=Π(2*1/P(n)^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/lnP(n)^k))*lnP(n)/2)

|ζ(x+i*y')-ζ(x+i*y)|=Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))*ln2/2) +Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))*ln3/2)
+Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))*ln4/2) ←　　Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2)＝０の時０に収束する

Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2) =2^a*1/2^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))*ln2/2) ←0に収束する必要がある
Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2) =2^a*1/3^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))*ln3/2) ←0に収束する必要がある
Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2) =2^a*1/4^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))*ln4/2) ←0に収束する必要がある

|ζ(x+i*y')-ζ(x+i*y)|=lim ΣΠ2^a*1/P(l)^(x*(1/2)/(1/2-1/2^∞))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/lnP(l)^k))*lnP(l)/2)=0
これが収束するときにx=1/2しかない可能性がある

http://rio2016.5ch.net/test/read.cgi/math/1640355175/312

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