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F(2n-1) = F(n-1)² + F(n)²
F(2n) = ( 2 F(n-1) + F(n) ) F(n)
Originally posted by dbates
No, we're not talking about generating Fibonacci numbers, this is about factoring Fibonacci numbers. You know, what two Fibonacci numbers make up a given Fibonacci number. For instance if you have x number of rabbits (where x is a Fibonacci number) that multiply every day, how many rabbits did we have 9 days ago. Well we had y males and z females. We know this by factoring the end result number.
Originally posted by sisonek
slank:
your formula only makes use of integers. you were talking about using it as a way of calculating the nth fibonacci, which gave you one formula for evens and one for odds, but there's a generating formula (see my earlier post) that gives you the n'th fibonacci in a straightforward manner. the catch, though, is that it involves the difference of two non-integer terms -- (1 + 5^(1/2)/2 and (1-5^(1/2))/2 -- to the n+1th power, so if you were to use it to calculate fibonacci's on a computer you'd either have to be in a computer algebra system that can handle the roots symbolically or you'd have to use floating points to do your calculation.
Nox:
just to pick nits, phi the "golden mean" isn't transcendental, just algebraic. algebraic = expressible as a root of a finite-degree polynomial with integer coefficients (like square root of two is a root of x^2 -2), transcendental = not expressible as a root of a finite-degree polynomial with integer coefficients (e and pi are famous transcendentals).