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F(2n-1) = F(n-1)˛ + F(n)˛
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
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.
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).