let f(n) be the n-th Fibonacci number (with f(0)=1, f(1)=1, f(n+2) = f(n+1)+f(n)):
The only number that can be represented as f(n)*f(f(n)^f(n+1))=f(1+f(n)^f(n+1))+f(f(n)*f(n+1))
I'll make it easier for you:
There was an open question which i have asked many proffesors and fellow mathematicians, but none were able to answer:
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)