Fibonacci Number Question?

Question

Is there any way of proving whether or not a number is a Fibonacci number other than looking through the list of numbers?

Answer 1

The nth Fibonacci number is:
(1+sqr(5))^n-(1-sqr(5))^n
-------------------------------
sqr(5)*2^n

Also, there are quite a few identities and relationships which may make it possible to prove by inspection that a number is NOT a Fibonacci number, but to prove that one is you have to find the correct n in the upper equation. For example, every 3rd number is even, every 6th number is divisible by 4, and every 25th is divisible by 25.

for more info read....http://mathworld.wolfram.com/FibonacciNumber.html.

Answer 2

Curt Monash is correct: There are many Fibonacci sequences. The "most famous one" he refers to has as its n'th term

(1/√5)[((1+√5)/2)^n - ((1-√5)/2)^n]

Unbelievable, isn't it! But try n = 1, 2, 3, 4 and you'll find it does give the terms
1, 1, 2, 3

So to find out whether a particular number, e.g. 9418, is a member of the sequence, you could put the above expression equal to it and try solving for n, to see if you get an integer value!!! But I don't know how such an equation could be solved.

Answer 3

I'm not optimistic.

First of all, I think one would have heard of it.

Second, there are many Fibonacci sequences; you're obviously just referring to the most famous of them, namely the one that is seeded with (1,1).

Answer 4

No, there is no way to tell unless you count up! I tried the same question for my science fair project!!!Hope this Helps!!!

Answer 5

this is really cool because last night for no particular reason i looked up the golden mean...i guess i can't really answer your question but i have witnessed some really cool syncronicity lately

Answer 6

No.