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

---

215: 2023/12/12(火)21:36 ID:mrhhK5hW(1/3) AAS
cos(2pi*(11^2/(2*3*5*7)^2))>cos(2pi*((2*a+1)/2^2+(3*b+1)/3^2+(5*c+1)/5^2+(d)/7^2)) > cos(2pi*(13*11/(2*3*5*7)^2))
a = 2 n_1, b = 3 n_2, c = 5 n_3 + 4, d = 49 n_4 + 39, n_1 element Z, n_2 element Z, n_3 element Z, n_4 element Z
a = 2 n_1 + 1, b = 3 n_2, c = 5 n_3, d = 49 n_4 + 5, n_1 element Z, n_2 element Z, n_3 element Z, n_4 element Z
cos(2pi*((2*2+1)/2^2+(3*3+1)/3^2+(5*4+1)/5^2+(39)/7^2)) =cos((131 π)/22050)
cos(2pi*((2*1+1)/2^2+(3*3+1)/3^2+(5*5+1)/5^2+(5)/7^2)) =cos((139 π)/22050)

---

216: 2023/12/12(火)22:56 ID:mrhhK5hW(2/3) AAS
cos(2pi*(11^2/(2*3*5*7)^2))>cos(2pi*((2*a+1)/2^2+(3*b+2)/3^2+(5*c+4)/5^2+(d)/7^2)) > cos(2pi*(13*11/(2*3*5*7)^2))　
a = 2 n_1, b = 3 n_2 + 2, c = 5 n_3 + 4, d = 49 n_4 + 44, n_1 element Z, n_2 element Z, n_3 element Z, n_4 element Z
a = 2 n_1 + 1, b = 3 n_2 + 2, c = 5 n_3, d = 49 n_4 + 10, n_1 element Z, n_2 element Z, n_3 element Z, n_4 element Z

e^(i*2pi*((2*2+1)/2^2+(3*2+1)/3^2+(5*4+1)/5^2+(44)/7^2)) =e^(-(10331 i π)/22050)
e^(i*2pi*((2*1+1)/2^2+(3*2+1)/3^2+(5*5+1)/5^2+(10)/7^2)) =e^(-(10061 i π)/22050)

---

217: 2023/12/12(火)23:54 ID:mrhhK5hW(3/3) AAS
e^(i*2pi*((2*2^n+1)/2^2+(3*2^n+2)/3^2+(5*2^n+4)/5^2+(7*2^n+8)/7^2))　
e^(i*2pi*((2*2^1+1)/2^2+(3*2^1+2)/3^2+(5*2^1+4)/5^2+(7*2^1+8)/7^2))=e^(-(1249 i π)/22050)
e^(i*2pi*((2*2+1)/2^2+(3*2+2)/3^2+(5*2+4)/5^2+(7*2+8)/7^2))＝e^((6521 i π)/22050)
e^(i*2pi*((2*4+1)/2^2+(3*4+2)/3^2+(5*4+4)/5^2+(7*4+8)/7^2))＝e^(-(22039 i π)/22050)
e^(i*2pi*((2*8+1)/2^2+(3*8+2)/3^2+(5*8+4)/5^2+(7*8+8)/7^2))＝e^((9041 i π)/22050)

---

上下前次1-新書関写板覧索設栞歴

---

ｽﾚ情報 赤ﾚｽ抽出 画像ﾚｽ抽出 歴の未読ｽﾚ AAｻﾑﾈｲﾙ

---

ぬこの手 ぬこTOP 0.022s