Recent content by Zerodivisor

You probably figured out to show f0*f1*...*fn = 2^2^n - 1 by expressing the terms in binary. So f0 = 3 = 11 binary, f1 = 101 binary, f0*f1 = 1100 + 11 = 1111 binary = 2^2^2 - 1, f0*f1*f2 = 1111 * 10001 = 11110000 + 1111 = 11111111 bin = 2^2^3 - 1, It's pretty how there are just enough zeros to...

OK, my error. The correct equation is (Concrete Mathematics, page 501): fn = 2 + f0*f1*...*fm*...*f(n-1) = 2 + k * fm so that fn % fm = 2. This is in fact the simple way to prove the fractional expression is 2.

I do not get this, Fy = 2 * prod(k=0 to y, Fk), where Fk = 2^(2^k))-1 , so Fy is relatively prime to all Fermat numbers less than it. qed