Prime counting function p(2^2)=4-4*(1/2)+mod(4,2)/2+1-1=2 : traditional sieve of prime can transform into p(4)=4*(1/2)+0/2=2, p(5)=5*(1/2)+1/2=3, p(6)=6*(1/2)+0/2=3, p(7)=7*(1/2)+1/2=4, p(3^2-1)=8*(1/2)+0/2=4, start at 3^2=9 add sieve of 3 have p(9)=9*(1/3)+1/2-3/6+0/3+1=4.......to p(24)=24*(1/3)+0/2-0/6+0/3+1=9, 1/3=(2-1)*(3-1)/(2*3) : partial reciprocal of Euler produce, start at 5^2=25 add sieve of 5, have (2-1)*(3-1)*(5-1)/(2*3*5)=4/15, p(2*3*5)=30*(4/15)+(0/2-0/6-0/10+0/30)+(0/3-0/15)+(0/5)+2=10, p(31)=31*(4/15)+(1-4/15)+2=11, 1-4/15=11/15=4/15+4/15+3/15 : sum of zero as Sir Michael Atiyak claimed F(s)=1-T(s)=1-ll(p-1)/p, (4/15)/(2-1)=4/15=1/2-1/6-1/10+1/30 : 1st zero correspond with 14.134... at large number x of p(x), (3-1)*(5-1)/(3*5)/(3-1)=4/15=1/3-1/15 : 2nd zero [21.02....], (5-1)/5/(5-1)=1/5=3/15 : 3rd zero[25.01...at long term], 4/15=4/15 to F(s)=2*F(s) Atiyah misunderstand 8/15=4/15 because every zero contain following zero show pattern 2^3=2^2+2^1+2^0+1: AGI of Hinton relate to Hogfield neural net work n!/(n-4)!*4!=n*(n-1)*(n^2-5n+18)/4! - n*(n-1)/2 relate to (n+m)!/(n!)*(m!) of Pascal triangle each row can deduce (1-2)-1=-2, (1-3)-(3-1)=-4, (1-4)-(6-4+1)=-6....-2n : trivial zero of zeta function to bottom of triangle contain nontrivial zero information, (2n)!/(n!)^2[1,2,6,20,70...] central line of Pascal triangle of Hilbert Polya random matrix divided by central line evenly prove RH by every zero start at pn^2[2^2, 3^2, 5^2....] x^(1/2)=e^(1/2*logx)=pn when x=pn^2, at infinite Euler prime x=(2*3*5*...*pn)+1 will all nontrivial zero on line 1/2 of x axis in complex plane because we have pattern can follow[zero start at pn^2] can use induction prove RH.
I do want to listen to this. Clicked on it so I don’t lose it.
I follow this subject.
Nice job everyone!!
Prime counting function p(2^2)=4-4*(1/2)+mod(4,2)/2+1-1=2 : traditional sieve of prime can transform into p(4)=4*(1/2)+0/2=2, p(5)=5*(1/2)+1/2=3, p(6)=6*(1/2)+0/2=3, p(7)=7*(1/2)+1/2=4, p(3^2-1)=8*(1/2)+0/2=4, start at 3^2=9 add sieve of 3 have p(9)=9*(1/3)+1/2-3/6+0/3+1=4.......to p(24)=24*(1/3)+0/2-0/6+0/3+1=9, 1/3=(2-1)*(3-1)/(2*3) : partial reciprocal of Euler produce, start at 5^2=25 add sieve of 5, have (2-1)*(3-1)*(5-1)/(2*3*5)=4/15, p(2*3*5)=30*(4/15)+(0/2-0/6-0/10+0/30)+(0/3-0/15)+(0/5)+2=10, p(31)=31*(4/15)+(1-4/15)+2=11, 1-4/15=11/15=4/15+4/15+3/15 : sum of zero as Sir Michael Atiyak claimed F(s)=1-T(s)=1-ll(p-1)/p, (4/15)/(2-1)=4/15=1/2-1/6-1/10+1/30 : 1st zero correspond with 14.134... at large number x of p(x), (3-1)*(5-1)/(3*5)/(3-1)=4/15=1/3-1/15 : 2nd zero [21.02....], (5-1)/5/(5-1)=1/5=3/15 : 3rd zero[25.01...at long term], 4/15=4/15 to F(s)=2*F(s) Atiyah misunderstand 8/15=4/15 because every zero contain following zero show pattern 2^3=2^2+2^1+2^0+1: AGI of Hinton relate to Hogfield neural net work n!/(n-4)!*4!=n*(n-1)*(n^2-5n+18)/4! - n*(n-1)/2 relate to (n+m)!/(n!)*(m!) of Pascal triangle each row can deduce (1-2)-1=-2, (1-3)-(3-1)=-4, (1-4)-(6-4+1)=-6....-2n : trivial zero of zeta function to bottom of triangle contain nontrivial zero information, (2n)!/(n!)^2[1,2,6,20,70...] central line of Pascal triangle of Hilbert Polya random matrix divided by central line evenly prove RH by every zero start at pn^2[2^2, 3^2, 5^2....] x^(1/2)=e^(1/2*logx)=pn when x=pn^2, at infinite Euler prime x=(2*3*5*...*pn)+1 will all nontrivial zero on line 1/2 of x axis in complex plane because we have pattern can follow[zero start at pn^2] can use induction prove RH.