Fibonacci number

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In mathematics, the Fibonacci numbers form a sequence defined by the following recurrence relation:

F_{n}:={\begin{cases}0&{\mbox{if }}n=0;\\1&{\mbox{if }}n=1;\\F_{n-1}+F_{n-2}&{\mbox{if }}n>1.\\\end{cases}}

The sequence of fibonacci numbers start: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

Fibonacci numbers and the rabbits

The sequence of fibonacci numbers was first used, to repesent the growth of a colony of rabbits, starting with one pair of rabbits.

The quotient of two consecutive fibonacci numbers converges to the golden ratio: $\lim _{n\to \infty }{\frac {F(n+1)}{F(n)}}=\varphi$
If $\scriptstyle m>2\$ divides $\scriptstyle n\$ then $\scriptstyle F_{m}\$ divides $\scriptstyle F_{n}\$
If $\scriptstyle p\geq 5$ is a prime number, then is $\scriptstyle F_{p}\$ also a prime number.
$\operatorname {gcd} (f_{m},f_{n})=f_{\operatorname {gcd} (m,n)}$
$F_{0}^{2}+F_{1}^{2}+F_{2}^{2}+...+F_{n}^{2}=\sum _{i=0}^{n}F_{i}^{2}=F_{n}\cdot F_{n+1}$

Let $A:={\frac {1+{\sqrt {5}}}{2}}$ and $a:={\frac {1-{\sqrt {5}}}{2}}$ . Let

f_{n}\ :=\ {\frac {1}{\sqrt {5}}}\cdot (A^{n}-a^{n})

Then:

for every $\ n=0,1,\dots$ . Thus $\ f_{n}=F_{n}$ for every $\ n=0,1,\dots$ , i.e.

 $F_{n}\ =\ {\frac {1}{\sqrt {5}}}\cdot \left(\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right)$

for every $\ n=0,1,\dots$ . Furthermore:

$A\cdot a=-1\$
$A>1\$
$-1<a<0\$
${\frac {1}{2}}\ >\ \left|{\frac {1}{\sqrt {5}}}\cdot a^{n}\right|\quad \rightarrow \quad 0$

It follows that

 $F_{n}\$   is the nearest integer to   ${\frac {1}{\sqrt {5}}}\cdot \left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}$

for every $\ n=0,1,\dots$ .