Transwiki:Fibonacci number

In mathematics, the Fibonacci numbers form a sequence defined recursively by:

F(n)= \left\{ \begin{matrix} 0\,,\qquad\qquad\qquad\quad\,\ \ \,&&\mbox{if }n=0\,;\ \ \\ 1,\qquad\qquad\qquad\qquad\,&&\mbox{if }n=1;\ \ \,\\ F(n-1)+F(n-2)&&\mbox{otherwise.} \end{matrix} \right. $$ In words: you start with 0 and 1, and then produce the next Fibonacci number by adding the two previous Fibonacci numbers. The first Fibonacci numbers (sequence A000045 in OEIS) for n = 0, 1,... are
 * 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946...

A tiling with Fibonacci number sized squares

Origins
This sequence was first described by Leonardo of Pisa, who was also known as Fibonacci (c. 1200), to describe the growth of a rabbit population. The numbers describe the number of pairs in a (somewhat idealized) rabbit population after n months if it is assumed that
 * the first month there is just one newly born pair,
 * newly born pairs become productive from their second month on,
 * we have no genetic problems whatsoever generated by inbreeding,
 * each month every productive pair begets a new pair, and
 * the rabbits never die

The formula above applies to the rabbit problem because if in month n we have a rabbits and in month n + 1 we have b rabbits then in month n + 2 we'll necessarily have a + b rabbits. That's because we know each rabbit basically gives birth to another each month (actually each pair gives birth to another pair, but it is the same thing) and that means that all a rabbits give birth to another number of a rabbits which will become fertile after two months, which is exactly at month n + 2. That's why we have the population at moment n + 1 (which is b) plus exactly the population at moment n (which is a).

The term Fibonacci sequence is also applied more generally to any function g where g(n + 2) = g(n) + g(n + 1). These functions are precisely those of the form g(n) = aF(n) + bF(n + 1) for some numbers a and b, so the Fibonacci sequences form a vector space with the functions F(n) and F(n + 1) as a basis.

In particular, the Fibonacci sequence L with L(1) = 1 and L(2) = 3 is referred to as the Lucas numbers. This sequence was described by Leonhard Euler in 1748, in the Introductio in Analysin Infinitorum. The significance in the Lucas numbers L(n) lies in the fact that raising the Golden ratio to the nth power yields:


 * $$\left( \frac 1 2 \left( 1 + \sqrt{5} \right) \right)^n = \frac 1 2 \left( L(n) + F(n) \sqrt{5} \right) $$

Lucas numbers are related to Fibonacci numbers by the relation:


 * $$L(n) = F(n-1) + F(n+1)$$

Explicit formula
As was pointed out by Johannes Kepler, the ratio of adjacent Fibonacci numbers, i.e. $$F(n+1)/F(n)$$, converges to the golden mean &phi;. In modern form:
 * $$x=\lim_{n\to\infty}\frac{F(n+1)}{F(n)}$$$$

=\lim_{n\to\infty}\frac{F(n)+F(n-1)}{F(n)}$$$$ =1+\lim_{n\to\infty}\frac{F(n-1)}{F(n)}$$$$ =1+\frac1{\lim_{n\to\infty}\frac{F(n)}{F(n-1)}}$$$$ =1+\frac1x$$ or $$x^2=x+1$$, a quadratic equation with $$\phi, 1-\phi$$ as solutions. If we multiply both sides by $$x^n$$, we get $$x^{n+2}=x^{n+1}+x^n$$. Thus the two functions $$n\mapsto\phi^n$$ and $$n\mapsto(1-\phi)^n$$ &mdash; as well as all linear combinations $$n\mapsto a\phi^n+b(1-\phi)^n$$ &mdash; satisfy the Fibonacci recurrence.

By adjusting the coefficients to get the proper initial values F(0) = 0 and F(1) = 1, we obtain


 * $$F\left(n\right) = {\phi^n \over \sqrt{5}} - {(1-\phi)^n \over \sqrt{5}}$$

This genrerates the closed form formula for the Fibonacci numbers as


 * $$F\left(n\right) = { (1+\sqrt{5})^n-(1-\sqrt{5})^n \over \sqrt{5}\cdot2^n }$$

This result can also be derived using the technique of generating functions, or the technique of solving linear recurrence relations.

As n goes to infinity, the second term converges to zero, so the Fibonacci numbers approach the exponential &phi;n/&radic;5, hence their convergent ratios. In fact the second term starts out small enough that the Fibonacci numbers can be obtained from the first term alone, by rounding to the nearest integer.

Computing Fibonacci numbers
Computing Fibonacci numbers by computing powers of the golden mean is not very practical except for small values of n, since rounding errors will accrue and floating point numbers usually don't have enough precision.

The straightforward recursive implementation of the Fibonacci sequence definition is also not advisable, since it would compute many values repeatedly (unless the programming language has a feature which allows the storing of previously computed function values). Therefore, one usually computes the Fibonacci numbers "from the bottom up", starting with the two values 0 and 1, and then repeatedly replacing the first number by the second, and the second number by the sum of the two.

For huge arguments and if a bignum system is being used, a faster way to calculate Fibonacci numbers uses the following matrix equation:


 * $$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n =

\begin{bmatrix} F\left(n+1\right) & F \left(n\right) \\ F\left(n\right)  & F \left(n-1\right) \end{bmatrix} $$

and employs exponentiating by squaring.

Applications
The Fibonacci numbers are important in the run-time analysis of Euclid's algorithm to determine the greatest common divisor of two integers.

Matiyasevich was able to show that the Fibonacci numbers can be defined by a Diophantine equation, which led to his original solution of Hilbert's tenth problem.

The Fibonacci numbers occur in a formula about the diagonals of Pascal's triangle (see binomial coefficient).

In music Fibonacci numbers are sometimes used to determine tunings, and, as in visual art, to determine the length or size of content or formal elements. Examples include Béla Bartók's Music for Strings, Percussion, and Celesta.

Generalizations
A generalization of the Fibonacci sequence are the Lucas sequences. One kind can be defined thus:


 * U(0) = 0
 * U(1) = 1
 * U(n+2) = PU(n+1) &minus; QU(n)

where the normal Fibonacci sequence is the special case of P = 1 and Q = -1. Another kind of Lucas Sequence begins with V(0) = 2, V(1) = P. Such sequences have applications in number theory and primality proving.

The Fibonacci polynomials are another generalization of Fibonacci numbers.

Identities

 * F(n + 1) = F(n) + F(n &minus; 1)


 * F(0) + F(1) + F(2) + ... + F(n) = F(n + 2) &minus; 1


 * F(1) + 2 F(2) + 3 F(3) + ... + n F(n) = n F(n + 2) &minus; F(n + 3) + 2

These identities can be proven using many different methods. But, among all, we wish to present an elegant proof for each of them using combinatorial arguments here. In particular, F(n) can be interpreted as the number of ways summing 1's and 2's to n &minus; 1, with the convention that F(0) = 0, meaning no sum will add up to -1, and that F(1) = 1, meaning the empty sum will "add up" to 0. Here the order of the summands matters. For example, 1 + 2 and 2 + 1 are considered two different sums and are counted twice.

Proof of the first identity. Without loss of generality, we may assume n &ge; 1. Then F(n + 1) counts the number of ways summing 1's and 2's to n.

When the first summand is 1, there are F(n) ways to complete the counting for n &minus; 1; and the first summand is 2, there are F(n &minus; 1) ways to complete the counting for n &minus; 2. Thus, in total, there are F(n) + F(n &minus; 1) ways to complete the counting for n. $$\Box$$

Proof of the second identity. We count the number of ways summing 1's and 2's to n + 1 such that at least one of the summands is 2.

As before, there are F(n + 2) ways summing 1's and 2's to n + 1 when n &ge; 0. Since there is only one sum of n + 1 that does not use any 2, namely 1 + &hellip; + 1 (n + 1 terms), we subtract 1 from F(n + 2).

Equivalently, we can consider the first occurrence of 2 as a summand. If, in a sum, the first summand is 2, then there are F(n) ways to the complete the counting for n &minus; 1. If the second summand is 2 but the first is 1, then there are F(n &minus; 1) ways to complete the counting for n &minus; 2. Proceed in this fashion. Eventually we consider the (n + 1)-th summand. If it is 2 but all of the previous n summands are 1's, then there are F(0) ways to complete the counting for 0. If a sum contains 2 as a summand, the first occurrence of such summand must take place in between the first and (n + 1)-th position. Thus F(n) + F(n &minus; 1) + &hellip; + F(0) gives the desired counting. $$\Box$$

Proof of the third identity. This identity can be establshed in two stages. First, we count the number of ways summing 1's and 2's to -1, 0, &hellip;, or n + 1 such that at least one of the summands is 2.

By our second identity, there are F(n + 2) &minus; 1 ways summing to n + 1; F(n + 1) &minus; 1 ways summing to n; &hellip;; and, eventually, F(2) &minus; 1 way summing to 1. As F(1) &minus; 1 = F(0) = 0, we can add up all n + 1 sums and apply the second identity again to obtain
 * [F(n + 2) &minus; 1] + [F(n + 1) &minus; 1] + &hellip; + [F(2) &minus; 1]
 * = [F(n + 2) &minus; 1] + [F(n + 1) &minus; 1] + &hellip; + [F(2) &minus; 1] + [F(1) &minus; 1] + F(0)
 * = F(n + 2) + [F(n + 1) + &hellip; + F(1) + F(0)] &minus; (n + 2)
 * = F(n + 2) + F(n + 3) &minus; (n + 2).

On the other hand, we observe from the second identity that there are &hellip;&hellip; Adding up all n + 1 sums, we see that there are
 * F(0) + F(1) + &hellip; + F(n &minus; 1) + F(n) ways summing to n + 1;
 * F(0) + F(1) + &hellip; + F(n &minus; 1) ways summing to n;
 * F(0) way summing to -1.
 * (n + 1) F(0) + n F(1) + &hellip; + F(n) ways summing to -1, 0, &hellip;, or n + 1.

Since the two methods of counting refer to the same number, we have
 * (n + 1) F(0) + n F(1) + &hellip; + F(n) = F(n + 2) + F(n + 3) &minus; (n + 2)

Finally, we complete the proof by subtracting the above identity from n + 1 times the second identity. $$\Box$$

Tribonacci number
A tribonacci number is like a Fibonacci number, but instead of starting with two predetermined terms, the sequence starts with three predetermined terms and each term afterwards is the sum of the preceding three terms. The first few tribonacci numbers are:

1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012

OEIS number: A000073

Tetranacci number
A tetranacci number is like a tribonacci number, only that starting with four predetermined terms and each term afterwards being the sum of the preceding four terms. The first few tetranacci numbers are

1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536, 10671, 20569, 39648, 76424, 147312, 283953, 547337

OEIS number: A000078

Pentanacci, hexanacci and heptanacci numbers have been computed, but they have not interested researchers much.

Repfigits
A repfigit or Keith number is an integer, that when its digits start a Fibonacci sequence with that number of digits, the original number is eventually reached. An example is 47, because the Fibonacci sequence starting with 4 and 7 (4,7,11,18,29,47) reaches 47. A repfigit can be a tribonacci sequence if there are 3 digits in the number, a tetranacci number if the number has four digits, etc. The first few repfigits are

14, 19, 28, 47, 61, 75, 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385, 7647, 7909

OEIS number: A007629