素数の規則を見つけたい。。。 (701ﾚｽ)
上下前次1-新
抽出解除 必死ﾁｪｯｶｰ(本家) (べ) 自ID ﾚｽ栞 あぼーん

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180: 2023/09/17(日)00:19 ID:NvL18fxN(1/3) AAS
e^(2 i π (2/13(2/11 (2/7 (2/5 (n/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2))

cos(2 π (2/13(2/11 (2/7 (2/5 (n/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2)) < cos(2π*17^2/(2*3*5*7*11*13))

1/16 (15015 m - 9101)<n<1/16 (15015 m - 8812)

e^(2 i π (2/13(2/11 (2/7 (2/5 (-551/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2))=e^((281 i π)/15015)
e^(2 i π (2/13(2/11 (2/7 (2/5 (-553/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2))=e^((217 i π)/15015) ←非素数
e^(2 i π (2/13(2/11 (2/7 (2/5 (-554/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2))=e^((185 i π)/15015) ←非素数
e^(2 i π (2/13(2/11 (2/7 (2/5 (-556/3 + 1/2) + 1/2) + 1/2) + 1/2)+1/2))=e^((121 i π)/15015) ←非素数
省8

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181: 2023/09/17(日)00:45 ID:NvL18fxN(2/3) AAS
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)
省3

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182: 2023/09/17(日)00:49 ID:NvL18fxN(3/3) AAS
連続する素数の差分は2^nと2^(n-1)が交互に来る
73 +2^4=89
89+2^3=97
97+2^4=113

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