The nature of the Fibonacci series

Fibonacci: 0, 1, 1, 2, 3, 5, 8, 13,... f [0] = 0;

1: gcd (FN, FM) = F [gcd (n, m)]; it can be proved by mathematical induction when n-M = 1 or 2;

2: The feature equation is x ^ 2 = x + 1, and the feature equation of the class of the Fibonacci series is: ax ^ 2 = bx + c; AF [N] = BF [n-1] + CF [n-2];

3: (the proof method is supplementary and mathematical induction)

F [0] + F [1] +... + F [N] = f [n + 2]-1;

F [0] + F [2] +... + F [2n] = f [2n + 1]-1;

F [1] + F [3] +... + F [2n-1] = f [2n];

F [0] ^ 2 + F [1] ^ 2 +... f [N] ^ 2 = f [N] * f [n + 1];

F [N] ^ 2 = (-1) ^ (n + 1) + F [n-1] * f [n + 1];

F [2n] = f [N] * (F [n + 1] + F [n-1]);

4: F [N] % x = 0 then f [N * k] % x = 0; k is an integer;

5: lim n-> oo f [n + 1]/f [N] = 0. 618..., prove that the method is the recursive formula, take the ratio on both sides, and then obtain the limit.

6: the N + 2 of the Fibonacci series also represents the set {1, 2 ,..., n} does not contain the number of subsets of adjacent positive integers. It is proved that, considering the N number, F [N] = f [n-1] + F [n-2], the boundary is determined by F [1] = 2;

7: relationship with the number of combinations: F (n) = C (n-1, 0) + C (n-2, 1) +... + C (n-1-m, m) (M <= n-1-m), sums the oblique diagonal corner of the Yang Hui triangle, and forms a Fibonacci Series

8: For the prime number P, F [N] % P has a cyclic section. If 5 is the quadratic residue of the modulo p, the length of the cyclic section is a factor of p-1, otherwise, it is a factor of 2 (p + 1), and the same is true for class Fibonacci;