Fibonacci number

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, ...

The sequence of Fibonacci numbers was first used to represent the growth of a colony of rabbits, starting with a single pair of rabbits.

Properties

We will apply the following simple observation to Fibonacci numbers:

if three integers $\ a,b,c,$  satisfy equality $\ c=a+b,$  then

$\ \gcd(a,b)\ =\ \gcd(a,c)=\gcd(b,c).$

• $\gcd \left(F_{n},F_{n+1}\right)\ =\ \gcd \left(F_{n},F_{n+2}\right)\ =\ 1$

Indeed,

$\gcd \left(F_{0},F_{1}\right)\ =\ \gcd \left(F_{0},F_{2}\right)\ =\ 1$

and the rest is an easy induction.

• $F_{n}\ =\ F_{k+1}\cdot F_{n-k}+F_{k}\cdot F_{n-k-1}$

for all integers $\ k,n,$  such that $\ 0\leq k<n.$

Indeed, the equality holds for $\ k=0,$  and the rest is a routine induction on $\ k.$

Next, since $\gcd \left(F_{k},F_{k+1}\right)=1$,  the above equality implies:

$\gcd \left(F_{k},F_{n}\right)\ =\ \gcd \left(F_{k},F_{n-k}\right)$

which, via Euclid algorithm, leads to:

• $\gcd(F_{m},F_{n})\ =\ F_{\gcd(m,n)}$

Let's note the two instant corollaries of the above statement:

• If $\ m$  divides $\ n\$ then $\ F_{m}\$ divides Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ F_n\ }

• If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ F_p\ }   is a prime number then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ p}   is prime. (The converse is false.)

• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=0}^n F_i^2 = F_n \cdot F_{n+1}}

Direct formula and the golden ratio

We have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n=0,1,\dots} .

Indeed, let  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A := \frac{1+\sqrt{5}}{2}}   and  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a := \frac{1-\sqrt{5}}{2}} .  Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n\ :=\ \frac{1}{\sqrt{5}}\cdot(A^n - a^n)}

Then:

• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_0 = 0\ }     and     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ f_1 = 1}
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^2 = A+1\ }     hence     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A^{n+2} = A^{n+1}+A^n}
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^2 = a+1\ }     hence     Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{n+2} = a^{n+1}+a^n\ }
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{n+2}\ =\ f_{n+1}+f_n}

for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n=0,1,\dots} . Thus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ f_n = F_n}   for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n=0,1,\dots,} and the formula is proved.

Furthermore, we have:

• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\cdot a = -1\ }
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A > 1\ }
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1 < a < 0\ }
• Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}\ >\ \left|\frac{1}{\sqrt{5}}\cdot a^n\right|\quad\rightarrow\quad 0}

It follows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n\ }   is the nearest integer to  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{5}}\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n}

for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n=0,1,\dots} . The above constant Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ A}   is known as the famous golden ratio Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \Phi.}   Thus:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi\ =\ \lim_{n\to\infty}\frac{F(n+1)}{F(n)}\ =\ \frac{1+\sqrt{5}}{2}}

Further reading

• John H. Conway und Richard K. Guy, The Book of Numbers, ISBN 0-387-97993-X