Fibonacci number

In mathematics, the Fibonacci numbers form a sequence in which the first number in the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers. In mathematical terms, it is 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 starts: 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.

Divisibility 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 $\ F_{n}\$

• If $\ F_{p}\$  is a prime number different from 3, then $\ p$  is prime. (The converse is false.)

Algebraic identities

• $F_{n-1}\cdot F_{n+1}-F_{n}^{2}\ =\ (-1)^{n}\$     for n=1,2,...
• $\sum _{i=0}^{n}F_{i}^{2}=F_{n}\cdot F_{n+1}$

Direct formula and the golden ratio

We have

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

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

Indeed, 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:

• $f_{0}=0\$     and     $\ f_{1}=1$
• $A^{2}=A+1\$     hence     $\ A^{n+2}=A^{n+1}+A^{n}$
• $a^{2}=a+1\$     hence     $a^{n+2}=a^{n+1}+a^{n}\$
• $f_{n+2}\ =\ f_{n+1}+f_{n}$

for every $\ n=0,1,\dots$. Thus $\ f_{n}=F_{n}$  for every $\ n=0,1,\dots ,$ and the formula is proved.

Furthermore, we have:

• $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 $\ n=0,1,\dots$ . The above constant $\ A$  is known as the famous golden ratio $\ \Phi .$  Thus:

$\Phi \ =\ \lim _{n\to \infty }{\frac {F(n+1)}{F(n)}}\ =\ {\frac {1+{\sqrt {5}}}{2}}$

Further reading

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