- 双子素数が無限個あることの証明
31 :132人目の素数さん[sage]:2018/01/13(土) 04:17:57.17 ID:VYlDrPFt - P(n+1)=√(P(n)*・・・*7*5*3*2)*((x^2+1/2^2+1/3^2+1/5^2+・・・+1/P(n)^2+2*(x*(1/2*cos(y*log2)+1/3*cos(y*log3)+1/5*cos(y*log5)+・・・+1/P(n)*cos(y*logP(n)))+
1/2*(1/3*cos(y*log3/2)+1/5*cos(y*log5/2)+・・・+1/P(n)*cos(y*logP(n)/2))+・・・+1/P(n-1)*1/P(n)*cos(y*logP(n)/P(n-1))))^(1/4) 3=√2*(4^2+1/2^2+2*(4*1/2*cos(0*log2)))^(1/4) 5=√(3*2)*(4^2+1/2^2+1/3^2+2*(4*(1/2-1/3)-1/2*1/3))^(1/4) =√(√2*2)*(4^2+1/2^2+2*(4*1/2*cos(0*log2)))^(1/8)*(4^2+1/2^2+1/3^2+2*(4*(1/2-1/3)-1/2*1/3))^(1/4) 7=√(5*3*2)*(1^2+1/2^2+1/3^2+1/5^2+2*(1*(1/2+1/3-1/5)+1/2*(1/3-1/5)-1/3*1/5))^(1/4) 7=√(√(√2*2)*√2*2)*(4^2+1/2^2+2*(4*1/2*cos(0*log2)))^(1/16)*(4^2+1/2^2+1/3^2+2*(4*(1/2-1/3)-1/2*1/3))^(1/8)*(1^2+1/2^2+1/3^2+1/5^2+2*(1*(1/2+1/3-1/5)+1/2*(1/3-1/5)-1/3*1/5))^(1/4)
|