Koala

The continuation of Fibonacci is one of the most known continuations Mathématiques. It owes its name to the mathematician Italy N Leonardo Pisano, more known under the pseudonym of Fibonacci (1175 - 1250). In an entertaining problem posed in one of its works, the Liber Abaci , Fibonacci describes the growth of a rabbit population:

“Having a rabbit couple initially, how much couples is it obtained in twelve months if each couple generates every month a new couple as from the second month of its existence? ”

This problem is at the origin of the continuation whose $n\; \backslash ,$-ème term corresponds to the number of pairs of rabbits to the $n\; \backslash ,$-ème month. In this population (ideal), one supposes that:

• the first month, there is just an even young rabbits;
• the young rabbits are pubescent only as from the second month;
• each month, any pair likely to procreate generates indeed a new pair of young rabbits;
• the rabbits never die (thus the continuation of Fibonacci is strictly increasing).

Mathematical presentation

Let us note $\backslash \; mathcal\; \{F\}\; \_n$ the number of couples of rabbits in the month $n\; \backslash ,$. Until the end of the second month, the population is limited to a couple (what one notes: $\backslash \; mathcal\; \{F\}\; \_1=\; \backslash \; mathcal\; \{F\}\; \_2=1$). At the beginning of the third month, the rabbit couple has two months and it generates another rabbit couple. One notes then $\backslash \; mathcal\; \{F\}\; \_3=2$. We place maintaining in the month $n\; \backslash ,$ and seek to express what it will be two months later $(N\; +\; 2)\; \backslash ,$: $\backslash \; mathcal\; \{F\}\; \_\; \{n+2\}$ indicates the sum of the couples of rabbits in the months $n\; +\; 2\; \backslash ,$ and of the lately generated couples. However, do not generate in the month $(N\; +\; 2)\; \backslash ,$ that the pubescent couples two months before. One thus has:

$\backslash \; mathcal\; \{F\}\; \_\; \{n+2\}\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; +\; \backslash \; mathcal\; \{F\}\; \_n$.

We obtain the recurring form thus of the continuation of Fibonacci: each term of this continuation is the sum of the two preceding terms; to obtain each one of these two terms, it is necessary to make the sum of their preceding terms… and so on, until these two terms are the two initial terms, $\backslash \; mathcal\; \{F\}\; \_1$ and $\backslash \; mathcal\; \{F\}\; \_2$, which is known.

The first values of the numbers of Fibonacci, in the order ascending while starting with $n=0$, are thus given by:

0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,…

The terms of this continuation are called numbers of Fibonacci .

Functional expression

One wishes to establish a functional expression of the continuation of Fibonacci, i.e. an expression such as the calculation of the number of couples for a value of $n\; \backslash ,$ given does not presuppose the knowledge of any number of couples for any another value of $n\; \backslash ,$, which the formula of recurrence does not allow. As the continuation of Fibonacci is recurring of order two, one can write his characteristic equation. One obtains a quadratic equation:

$x^2-x-1=0\; \backslash ,$.

The calculation of the Discriminant of this equation gives the two solutions of the Polynôme:

$x\_1\; =\; \backslash \; varphi\; =\; \backslash \; frac\; \{1+\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash ,$ and $x\_2\; =\; \backslash \; varphi\text{'}\; =\; \backslash \; frac\; \{1\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash ,$. ($\backslash \; varphi\; \backslash ,$ is the Golden section, and $\backslash \; varphi\text{'}\; \backslash ,$ the opposite of the gilded section).

The continuations $(\backslash \; varphi^n)$ and $(\backslash \; varphi\text{'}^n)$ then generate the vector Space continuations checking $u\_\; \{n+2\}\; =u\_\; \{n+1\}\; +u\_n$. It results from it that:

$\backslash \; mathcal\; \{F\}\; \_n\; =\; \backslash \; alpha\; \{\}\; \backslash \; varphi^n+\; \backslash \; beta\; \backslash \; varphi\text{'}^n$ ($\backslash \; alpha\; \backslash ,$ and $\backslash \; beta\; \backslash ,$ is constants to be determined starting from $\backslash \; mathcal\; \{F\}\; \_1$ and $\backslash \; mathcal\; \{F\}\; \_2$.)

The initial conditions $\backslash \; mathcal\; \{F\}\; \_1=\; \backslash \; mathcal\; \{F\}\; \_2=1$ lead to the following system:

$\backslash \; left\; \backslash \; \{\backslash \; begin\; \{matrix\}\; \backslash \; alpha\; +\; \backslash \; beta\; =\; 0\; \backslash \; \backslash \; \backslash \; alpha\; -\; \backslash \; beta\; =\; \backslash \; frac\; \{2\}\; \{\backslash \; sqrt\; \{5\}\}\; \backslash \; end\; \{matrix\}\; \backslash \; right.$

What gives the following result:

$\backslash \; alpha\; =\; \backslash \; frac\; \{1\}\; \{\backslash \; sqrt\; \{5\}\}$ and $\backslash \; beta\; =\; -\; \backslash \; frac\; \{1\}\; \{\backslash \; sqrt\; \{5\}\}$.

We obtain finally the required functional expression, which bears the name of formula of Binet :

$\backslash \; mathcal\; \{F\}\; \_n\; =\; \backslash \; frac\; \{1\}\; \{\backslash \; sqrt\; \{5\}\}\; (\backslash \; varphi^n-\; \backslash \; varphi\text{'}^n)$.

There exist other demonstrations. A demonstration using the Transformée into Z is given in the article of the same name. The result is also obtained by using the technique of the generating functions.

The continuation for the negative numbers

In general, one does not study the numbers of Fibonacci for negative values of N, although they exist and are easily determinable with the recurring formula. There exists thus a very simple rule to generate these numbers when :

• if N is even then $\backslash \; mathcal\; \{F\}\; (-\; N)\; =\; -\; \backslash \; mathcal\; \{F\}\; (N)$
• if N is odd then $\backslash \; mathcal\; \{F\}\; (-\; N)\; =\; \backslash \; mathcal\; \{F\}\; (N)$

Limit of the quotients

As noticed it Johannes Kepler, the growth rate of the numbers of Fibonacci, i.e. $\backslash \; frac\; \{\backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\}\; \{\backslash \; mathcal\; \{F\}\; \_n\}$, converges towards the Golden section, noted $\backslash \; varphi\; \backslash ,$. Mathematically, the result is obtained as follows:

More precisely, when $n\; \backslash ,$ tends towards the infinite one, the second term tends towards zero because $\backslash \; varphi\text{'}\; \backslash \; in]\; -\; 1;\; 1\; thus\; the\; numbers\; of\; Fibonacci\; behave\; like\; exponential\; multiplied\; by\; the\; factor$ \backslash \; frac\; \{1\}\; \{\backslash \; sqrt\; \{5\}\}$,\; that\; is\; to\; say$ \backslash \; frac\; \{\backslash \; varphi^n\}\; \{\backslash \; sqrt\; \{5\}\}$.$

In fact, as of the row $n=1\; \backslash ,$, the second term $\{\backslash \; varphi\text{'}^n\; \backslash \; over\; \backslash \; sqrt\; \{5\}\}$ is rather small nearest so that the numbers of Fibonacci can be only obtained starting from the first term, while rounding with the entirety.

Bases and vector spaces

• the denomination of “continuation of Fibonacci generalized” is allotted more generally to all function $\backslash \; mathcal\; \{G\}$ definite on $\backslash \; mathbb\; N$ checking for entire naturalness $n\; \backslash ,$, $\backslash \; mathcal\; \{G\}\; \_\; \{n+2\}\; =\; \backslash \; mathcal\; \{G\}\; \_\; \{n+1\}\; +\; \backslash \; mathcal\; \{G\}\; \_n$. These functions are precisely those for which there exist numbers has and B , such as for entire naturalness N , $\backslash \; mathcal\; \{G\}\; \_n\; =\; has\; \{\}\; \backslash \; cdot\; \{\}\; \backslash \; mathcal\; \{F\}\; \_n+b\; \{\}\; \backslash \; cdot\; \{\}\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}$. Thus, the whole of the series Fibonacci is a vector Space, and the continuations $(\backslash \; mathcal\; \{F\}\; \_n)$ and $(\backslash \; mathcal\; \{F\}\; \_\; \{n+1\})$ form a Base of it.

• the Golden section is the positive root of the quadratic equation $x^2\; -\; X\; -\; 1\; =\; 0\; \backslash ,$, thus $\backslash \; varphi^2\; =\; \backslash \; varphi\; +\; 1\; \backslash ,$. If one multiplies the two sides by $\backslash \; varphi^n\; \backslash ,$, one obtains $\backslash \; varphi^\; \{n+2\}\; =\; \backslash \; varphi^\; \{n+1\}\; +\; \backslash \; varphi^n\; \backslash ,$, therefore the function $n\; \backslash \; mapsto\; \backslash \; varphi^n$ is a continuation of Fibonacci. The negative root of the quadratic equation, $\backslash \; varphi\text{'}=1-\; \backslash \; varphi\; \backslash ,$, has the same properties, and the two linearly independent functions $n\; \backslash \; mapsto\; \backslash \; varphi^n$ and $n\; \backslash \; mapsto\; \backslash \; left\; (1\; \backslash \; varphi\; \backslash \; right)\; ^n$, form another base of the vector space.

Calculation algorithms of the numbers of Fibonacci

With the Formula of Binet

To calculate the numbers of Fibonacci starting from the golden section is a very practical possibility. Nevertheless, the precision of calculation of the square root generates rounding errors for rather large values depending on the system used. In general, one obtains the good values until $\backslash \; mathcal\; \{F\}\; \_\; \{71\}\; =\; 308061521170130$, on computer or computer.

Let us note that beyond $\backslash \; mathcal\; \{F\}\; \_\; \{79\}$, calculations exceed the possibilities of calculation in whole notation, and are then represented in scientific notation. The first significant figures then again are well represented by this formula.

Detail of a feasible example of application starting from a Computer: Calculation of $\backslash \; mathcal\; \{F\}\; \_\; \{50\}$.

The golden section is worth: $\backslash \; frac\; \{1+\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash \; =\; 1,61803398874989$

Let us apply the formula of Binet, (by retaining that the significant term) that is to say:

$\backslash \; mathcal\; \{F\}\; \_\; \{50\}\; =\; \backslash \; frac\; \{(1,61803398874989)\; ^\; \{50\}\}\; \{\backslash \; sqrt\; \{5\}\}\; =\; 12586269024,9981$

Let us round with the entirety nearest is:

$\backslash \; mathcal\; \{F\}\; \_\; \{50\}\; =\; 12586269025$

Naive recursive algorithm

The recursive implementation naive which follows the definition of the continuation of Fibonacci is immediate. In C, that gives:

unsigned int fibo (unsigned int N) { yew (N < 2) return N; return fibo (n-1) + fibo (N2); }

It is however not a way judicious to calculate the continuation of Fibonacci, because one calculates many times the same values (unless employing a technique of Mémorisation). The Time computing proves Exponentiel.

Final recursive version: unsigned int fibo (unsigned int nb1, unsigned int nb2, unsigned int N) { yew (n==0) return 0; yew (n<2) return nb2; return fibo (nb2, nb1+nb2, n-1); }

Linear algorithm

An average more effective good to calculate the continuation of Fibonacci consists in simultaneously calculating two consecutive values of the continuation, i.e. while starting with the first two values 0 and 1, and by replacing répétitivement the first number by the second, and the second number by the sum of both.

In Java or C, that gives:

int F (int N) { int has; int B; int C; int I; = 0 have; B = 1; yew (N < 1) { return N; } else { I = 1; while (I < N) { C = has + B; = B has; B = C; I = I + 1; } } return C; }

In an equivalent way, one can write a final function recursive:

int fibo (int nb1, int nb2, int N) { yew (n==0) return 0; yew (n<2) return nb2; return fibo (nb2, nb1+nb2, n-1); }

int fibonacci (int N) { yew (n==0 || n==1) { return (N); } else { return (fibonacci (n-1) + fibonacci (N2)); } }

In:

function fibonacci ($n) { $a = 0; $b = 1; $c = 1; yew ($n == 0 || $n == 1) { return $n; } else { for ($i = 1; $i < $n; $i++) { $c = $a + $b; $a = $b; $b = $c; } return $c; } }

Or in Design, without loop: ; Use: (fibo N 0 1) (define (fibo N U v) (cond ((= N 0) U) ((= N 1) v) ((fibo (- N 1) v (+ U v)) )))

The computing time is each time proportional to N and it constant occupied memory capacity.

Algorithm logarithmic curve

With a system allowing arithmetic calculations into multiprecision, one more quickly calculates the numbers of Fibonacci for great values of the entirety N , by using the following matric relation, which one shows by recurrence:

\ begin {bmatrix} \ mathcal F_ {n+1} & \ mathcal F_n \ \ \ mathcal F_n & \ mathcal F_ {n-1} \ end {bmatrix}

and the fast algorithm of Exponentiation. The algorithm obtained is comparable with recursive programming using the following relations:

$\backslash \; mathcal\; F\_\; \{2k\}\; =\; \backslash \; mathcal\; F\_k^2\; +\; 2\; \backslash \; mathcal\; F\_k\; \backslash \; mathcal\; F\_\; \{k-1\}$

$\backslash \; mathcal\; F\_\; \{2k+1\}\; =\; \backslash \; mathcal\; F\_k^2\; +\; \backslash \; mathcal\; F\_\; \{k+1\}\; ^2$

The computing time is then proportional to the logarithm of N . Here a sample program in Maple ( iquo indicating the quotient in Euclidean division):

F: =proc (N) option remember; room m; yew n=0 then 0 elif n=1 then 1 else m: =iquo (N, 2): yew N MOD 2 = 0 then F (m) ^2+2*F (M-1) *F (m) else F (m) ^2+F (m+1) ^2 fi: fi: end:

One still gains in effectiveness by working over again the relations of recurrence to avoid redundant calculations:

$\backslash \; mathcal\; F\_\; \{2k\}\; =\; \backslash \; mathcal\; F\_\; \{2k+1\}\; -\; \backslash \; mathcal\; F\_\; \{2k-1\}\; =\; (\backslash \; mathcal\; F\_k^2\; +\; \backslash \; mathcal\; F\_\; \{k+1\}\; ^2)\; -\; (\backslash \; mathcal\; F\_\; \{k-1\}\; ^2\; +\; \backslash \; mathcal\; F\_k^2)\; =\; \backslash \; mathcal\; F\_\; \{k+1\}\; ^2\; -\; \backslash \; mathcal\; F\_\; \{k-1\}\; ^2$

One obtains finally for any positive index:

$\backslash \; mathcal\; F\_\; \{K\}\; =\; \backslash \; left\; \backslash \; \{\backslash \; begin\; \{matrix\}$

0 & \ mbox {if} k=0 \ \ 1 & \ mbox {if} 0

Here a sample program in Python: def F (N): yew N < 3: return 0 + (0 < N) elif N % 2: return F (N/2 + 1) ** 2 + F ((N - 1)/2) ** 2 else: return F (N/2 + 1) ** 2 - F ((N - 1)/2) ** 2

While extending to the negative indices one obtains:

$\backslash \; mathcal\; F\_\; \{K\}\; =\; \backslash \; left\; \backslash \; \{\backslash \; begin\; \{matrix\}$

-1 & \ mbox {if} k=-2 \ \ 1 & \ mbox {if} k=-1 \ mbox {or if} 0

Here a sample program in Python: def F (N): yew ABS (N) < 3: return - (N == -2) + (N == -1) + 0 + (0 < N) elif N % 2: return F (N/2 + 1) ** 2 + F ((N - 1)/2) ** 2 else: return F (N/2 + 1) ** 2 - F ((N - 1)/2) ** 2

Algorithmic curiosity

A way particularly curious to obtain the continuation of Fibonacci is the following one. One considers the list of fractions 57/23, 17/39, 130/17, 11/14, 35/11, 19/13, 1/19, 35/2, 13/7, 7. If one starts from an entirety of the form $2^\; \{F\; (n-1)\}3^\; \{F\; (N)\}$ and if one repeatedly multiplies it by the first fraction which gives again a whole result, then the first integer of the continuation thus obtained which will have only 2 and 3 as factors first will be the number $2^\; \{F\; (N)\}3^\; \{F\; (n+1)\}$.

For example, if one starts from $18\; =\; 2^1\; 3^2$, one obtains successively:

18 $\backslash \; times$ 35/2 = 315

315 $\backslash \; times$ 13/7 = 585

585 $\backslash \; times$ 17/39 = 255

255 $\backslash \; times$ 130/17 = 1950

1950 $\backslash \; times$ 17/39 = 850

850 $\backslash \; times$ 130/17 = 6500

6500 $\backslash \; times$ 19/13 = 9500

9500 $\backslash \; times$ 23/95 = 2300

2300 $\backslash \; times$ 57/23 = 5700

5700 $\backslash \; times$ 23/95 = 1380

1380 $\backslash \; times$ 57/23 = 3420

3420 $\backslash \; times$ 23/95 = 828

828 $\backslash \; times$ 57/23 = 2052

2052 $\backslash \; times$ 1/19 = 108

with 108 which is worth $2^2\; 3^3$. If the process is reiterated, one will see ravelling the numbers of Fibonacci in the exhibitors of the powers of 2 and 3 breaking up the numbers thus obtained.

Properties of the continuation of Fibonacci

The continuation of Fibonacci, thus defined, present of remarkable properties. Here are some, given with their demonstration (those call in general upon the Raisonnement by recurrence). We also give some properties binding the continuation of Fibonacci and the numbers of Lucas.

Property 1 : $\backslash \; forall\; (p,\; Q)\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^*\; \backslash \; times\; \backslash \; mathbb\; \{NR\},\; \backslash \; mathcal\; \{F\}\; \_\; \{p+q\}\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{p-1\}\; \backslash \; mathcal\; \{F\}\; \_q\; +\; \backslash \; mathcal\; \{F\}\; \_p\; \backslash \; mathcal\; \{F\}\; \_\; \{q+1\}$

Property 2 : $\backslash \; forall\; (K,\; N)\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \backslash \; times\; \backslash \; mathbb\; \{NR\}\; ^*,\; \backslash \; mathcal\; \{F\}\; \_n\; \{\}|\{\}\; \backslash \; mathcal\; \{F\}\; \_\; \{N\; \backslash \; cdot\; K\}$

Remark : One can draw an interesting conclusion from this property. Indeed, if $\backslash \; mathcal\; \{F\}\; \_n$ is first, then this one admits for dividers only the unit and itself (like their opposites). There thus does not exist entirety $m\; \backslash ,$ such as $\backslash \; mathcal\; \{F\}\; \_m\; \backslash \; not\; =\; 1$, $\backslash \; mathcal\; \{F\}\; \_m\; \backslash \; not\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{N\}$ and $\backslash \; mathcal\; \{F\}\; \_m\; \{\}|\{\}\; \backslash \; mathcal\; \{F\}\; \_\; \{N\}$, in other words: there does not exist entirety $m\; \backslash ,$ ($m\; \backslash \; not\; =\; 1$ and $m\; \backslash \; not\; =\; n$) which divides $n\; \backslash ,$, from where: $n\; \backslash ,$ is first. One notes:

$\backslash \; mathcal\; \{F\}\; \_n\; \backslash \; in\; \backslash \; mathbb\; \{P\}\; \{\}\; \backslash \; Rightarrow\; \{\}\; N\; \backslash \; in\; \backslash \; mathbb\; \{P\}$ where $\backslash \; mathbb\; \{P\}$ indicates the whole of the prime numbers (the reciprocal one is false: $2\; \backslash \; in\; \backslash \; mathbb\; \{P\},\; \backslash \; mathcal\; \{F\}\; \_2\; \{\}\; \backslash \; not\; \backslash \; in\; \{\}\; \backslash \; mathbb\; \{P\}$).

Property 3 : $\backslash \; forall\; (K,\; N)\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^2\; \{\}/\{\}\; N\; \{\}\; \backslash \; Ge\; \{\}\; K\; \{\}\; \backslash \; Ge\; \{\}\; 0,\; \backslash \; mathcal\; \{F\}\; \_n\; \{\}\; \backslash \; mathcal\; \{F\}\; \_\; \{k+1\}\; -\; \backslash \; mathcal\; \{F\}\; \_k\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; =\; (-\; 1)\; ^k\; \backslash \; mathcal\; \{F\}\; \_\; \{n-k\}$

Property 4 : $\backslash \; forall\; (p,\; Q)\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^2\; \{\},\; \backslash \; mathcal\; \{F\}\; \_p\; \{\}\; \backslash \; Land\; \{\}\; \backslash \; mathcal\; \{F\}\; \_q\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{p\; \{\}\; \backslash \; Land\; \{\}\; Q\}$

In particular, $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \{\},\; \backslash \; mathcal\; \{F\}\; \_n\; \{\}\; \backslash \; Land\; \{\}\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; =1$

Property 5 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^*\; \{\},\; \backslash \; mathcal\; \{L\}\; \_n\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}\; +\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}$

(Precision: the numbers of Lucas $\backslash \; mathcal\; \{L\}\; \_n$ have even relation of recurrence but have as an initialization $\backslash \; mathcal\; \{L\}\; \_0\; =\; 2$ and $\backslash \; mathcal\; \{L\}\; \_1\; =\; 1$)

Property 6 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; -\; \backslash \; \{0,1,2\; \backslash \}\; \{\},\; 2\; \backslash \; mathcal\; \{L\}\; \_n=\; \backslash \; mathcal\; \{F\}\; \_\; \{n-3\}\; +\; \backslash \; mathcal\; \{F\}\; \_\; \{n+3\}$

Property 7 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \{\},\; \backslash \; mathcal\; \{F\}\; \_\; \{2\; \{\}\; \backslash \; cdot\; \{\}\; N\}\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{N\}\; \backslash \; mathcal\; \{L\}\; \_\; \{N\}$

Property 8 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^*\; \{\},\; \backslash \; mathcal\; \{F\}\; \_\; \{2\; \{\}\; \backslash \; cdot\; \{\}\; n-1\}\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}\; ^2+\; \backslash \; mathcal\; \{F\}\; \_\; \{N\}\; ^2$

by direct application of property 1 for p = N and Q = N -1

Property 9 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^*\; \{\},\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}\; -\; \backslash \; mathcal\; \{F\}\; \_n^2=\; (-\; 1)\; ^n$

Property 10 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \{\},\; 1+\; \backslash \; sum\_\; \{i=0\}\; ^n\; \backslash \; mathcal\; \{F\}\; \_i=\; \backslash \; mathcal\; \{F\}\; \_\; \{n+2\}$

Property 11 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \{\},\; 1+\; \backslash \; sum\_\; \{i=0\}\; ^n\; \backslash \; mathcal\; \{F\}\; \_\; \{2\; \{\}\; \backslash \; cdot\; \{\}\; I\}\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{2\; \{\}\; \backslash \; cdot\; \{\}\; n+1\}$

Property 12 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; \{\},\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; =\; \backslash \; sum\_\; \{k=0\}\; ^\; \{\backslash \; infty\}\; \{n-k\; \backslash \; choose\; K\}$ where the $n-k\; \backslash \; choose\; k$ are binomial coefficients.

That means that, in Triangle of Pascal, the numbers of Fibonacci are obtained by summoning the terms located on a diagonal (of bottom towards the line)

Property 13 : $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\}\; ^*,\; \backslash \; varphi^n\; =\; \backslash \; mathcal\; \{F\}\; \_n\; \{\}\; \backslash \; cdot\; \{\}\; \backslash \; varphi\; +\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}$

Property 14 : $\backslash \; forall\; N\; >\; 2,\; \backslash \; mathcal\; \{F\}\; \_\; \{n+2\}\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}\; \backslash \; mathcal\; \{F\}\; \_\; \{N2\}\; -\; \backslash \; mathcal\; \{F\}\; \_\; \{N\}\; ^4\; +\; 1\; =\; 0$

Formulas

• the following formula makes it possible to find all the numbers of Fibonacci (formula of Binet):

$\backslash \; mathcal\; \{F\}\; \_\; \{N\}\; =\; \backslash \; frac\; \{1\}\; \{\backslash \; sqrt\; \{5\}\}\; \backslash \; left\; \backslash \; left\; (\backslash \; frac\; \{1\; +\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash \; right)\; ^\; \{N\}\; -\; \backslash \; left\; (\backslash \; frac\; \{1\; -\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash \; right)\; ^\; \{N\}\; \backslash \; right$

• the following formula makes it possible to find all the numbers of Lucas (formula of Mallet):

$\backslash \; mathcal\; \{L\}\; \_\; \{N\}\; =\; \backslash \; left\; \backslash \; left\; (\backslash \; frac\; \{1\; +\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash \; right)\; ^\; \{N\}\; +\; \backslash \; left\; (\backslash \; frac\; \{1\; -\; \backslash \; sqrt\; \{5\}\}\; \{2\}\; \backslash \; right)\; ^\; \{N\}\; \backslash \; right$

Primality of the numbers of Fibonacci

One is unaware of if there exists an infinity of numbers of Fibonacci first. It is known that $\backslash \; mathcal\; \{F\}\; \_n$ divides $\backslash \; mathcal\; \{F\}\; \_\; \{K\; \{\}\; \backslash \; cdot\; \{\}\; N\}$ (see Propriétés, Propriété 2 ), and thus that, for all $n\; >4\; \backslash ,$, if $\backslash \; mathcal\; \{F\}\; \_n$ is first, then $n\; \backslash ,$ is first, but the reciprocal one is false. The greatest number of Fibonacci first known is $\backslash \; mathcal\; \{F\}\; \_\; \{604711\}$ which has 126377 digits.

There exist continuations $(\backslash \; mathcal\; \{T\}\; \_n)$ checking the three following conditions at the same time:

• $\backslash \; mathcal\; \{T\}\; \_\; \{n+2\}\; =\; \backslash \; mathcal\; \{T\}\; \_\; \{n+1\}\; +\; \backslash \; mathcal\; \{T\}\; \_n$
• $\backslash \; mathcal\; \{T\}\; \_n\; \{\}$ and $\backslash \; mathcal\; \{T\}\; \_\; \{n+1\}$ is first between them (they do not have any common Diviseur).
• $\backslash \; forall\; N\; \backslash \; in\; \backslash \; mathbb\; \{NR\},\; \backslash ;\; \backslash \; mathcal\; \{T\}\; \_n$ is not first.

The smallest known examples are determined by:

• $\backslash \; mathcal\; \{T\}\; \_0\; =\; 3794765361567513\; =\; 3\; \backslash \; cdot\; 1264921787189171$
• $\backslash \; mathcal\; \{T\}\; \_1\; =\; 20615674205555510\; =\; 2\; \backslash \; cdot\; 5\; \backslash \; cdot\; 5623\; \backslash \; cdot\; 366631232537$

• $\backslash \; mathcal\; \{T\}\; \_0\; =\; 1786772701928802632268715130455793\; =\; 2521\; \backslash \; cdot\; 49993\; \backslash \; cdot\; 14177095479037851751198481$

• $\backslash \; mathcal\; \{T\}\; \_1\; =\; 1059683225053915111058165141686995\; =\; 3\; \backslash \; cdot\; 5\; \backslash \; cdot\; 84089\; \backslash \; cdot\; 73919059\; \backslash \; cdot\; 150031897\; \backslash \; cdot\; 75754002239$

Applications

• the continuation of Fibonacci appears in many problems of enumeration. For example, the term of index N (for N equal to or higher than 2) of the continuation of Fibonacci makes it possible to count the number of ways of traversing a way length n-1 by taking steps of 1 or 2. This problem appears besides very early in India, under the name maatraameru (mountain of rate), in the work of the grammairien of Sanskrit Pingala, the Chhandah-shastra , (the art of the Prosody), 450 or 200 av. JC). The mathematician Indien Virahanka gave explicit rules of them to VIIIe century. The Indian philosopher Hemachandra (C. 1150) (and also Gopala) revisited the problem in a rather detailed way. In Sanskrit indeed, the vowels can be long (L) or short (C), and Hemachandra wished to calculate how much one can form rates different a given length, each rate being defined by the lengths of the vowels which constitute it. If the long vowel is twice longer than the short one, the solutions are, according to the overall length of rate:

1 C → --> 1

2 DC, L → --> 2

3 CCC, CL, LLC → --> 3

4 CCCC, CCL, CLC, LCC, L → --> 5

5 CCCCC, CCCL, CCLC, CLCC, LCCC, CL, LCL, LLC → --> 8

the number of rates reveals the terms of the continuation of Fibonacci. Indeed, a rate length N can be made up by adding C to a rate length n-1 , or L at a rate length N2 . Thus the number of rates length N is the sum of the two preceding numbers of the series. This problem is also equivalent to the enumeration of packing length N given, consisted of articles length 1 or 2, such as one off finds it for example in The Art Computer Programming of Donald Knuth.

• the numbers of Fibonacci intervene in the study of the execution of the Algorithme of Euclide which determines the highest common factor of two entireties.

• Matiyasevich showed that the numbers of Fibonacci could be defined by a equation diophantienne, which led to the resolution of the tenth problem of Hilbert. In 1975, Jones in deduced that, for values of X and there whole positive or null, the positive values of the polynomial $2xy^4\; +\; x^2y^3\; -\; 2x^3y^2\; -\; y^5\; -\; x^4y\; +\; 2y$ was exactly the numbers of Fibonacci. These positive values are obtained besides while allotting for values to X and there two numbers of Fibonacci successive.

• the numbers of Fibonacci appear in the formula of the diagonals of the Triangle of Pascal (see Propriétés, Propriété 11 ).

• an interesting use of the series Fibonacci is the conversion of the miles into kilometers. For example, to know how much kilometers make 5 miles, it is enough to consider the $\backslash \; mathcal\; \{F\}\; \_5\; =\; 5$ and following it $\backslash \; mathcal\; \{F\}\; \_6\; =\; 8$. 5 miles make approximately 8 kilometers. That functions because the conversion factor between the miles and the kilometers is coarsely equal to $\backslash \; varphi\; \backslash ,$.

• a good approximation of a right-angled gold can be built using squares whose sides are equal to the numbers of Fibonacci.

• a Spirale logarithmic curve can be in the following way approximate: one starts at the origin of a Cartesian reference mark, one moves $\backslash \; mathcal\; \{F\}\; \_1$ units towards the line, then of $\backslash \; mathcal\; \{F\}\; \_2$ units to the top, one moves $\backslash \; mathcal\; \{F\}\; \_3$ units towards the left, then of $\backslash \; mathcal\; \{F\}\; \_4$ units downwards, then of $\backslash \; mathcal\; \{F\}\; \_5$ units towards the line, etc That resembles the construction mentioned in the article on the Golden section. The numbers of Fibonacci often appear in nature when spirals logarithmic curves are built starting from a discrete unit, such as in the sunflowers or pine cones.

• the majority of the sexués living beings result from two parents, so that their ancestors with N E generation, presumedly distinct, are 2 ^N . But the hyménoptères are such as the females result from two parents, and the males result from a mother only. It results from it that their ancestors with N E generation are made up:

for the females, of $\backslash \; mathcal\; F\_n$ males and $\backslash \; mathcal\; female\; F\_\; \{n+1\}$,

for the males, of $\backslash \; mathcal\; male\; F\_\; \{n-1\}$ and $\backslash \; mathcal\; F\_n$ females.

• the number in different ways of pave a Rectangle 2×N by means of dominos 2×1 is $\backslash \; mathcal\; F\_\; \{N+1\}$.

Generalizations

There exist several ways to generalize the continuation of Fibonacci: one can modify the initial values, modify the coefficients of the relation of recurrence or modify the number of terms (or order) of the relation of recurrence.

Generalized series Fibonacci

In fact continuations preserve the same relation of recurrence but whose initial terms changed. As showed it the first part, these continuations are linear combinations of the two geometrical continuations $(\backslash \; varphi)\; ^n$ and $(1\; \backslash \; varphi)\; ^n$ where φ is the Golden section. The quotient of two consecutive terms always tends towards φ.

Among these continuations of numbers, it is necessary to announce the numbers of Lucas obtained while choosing like initialization: $L\_0\; =2$ and $L\_1=1$. That takes action pursuant 2,1,3,4,7,11,18,29,… One finds sometimes an initialization $L\_0\; =1$ and $L\_1=3$ which only consists in shifting the continuation of a row. These numbers intervene in the resolution of equations diophantiennes. They are very dependant following Fibonacci, in particular by the following relation: $\backslash \; mathcal\; \{L\}\; \_n\; =\; \backslash \; mathcal\; \{F\}\; \_\; \{n+1\}\; +\; \backslash \; mathcal\; \{F\}\; \_\; \{n-1\}\; \backslash ,$ for entire strictly positive N (see Properties, Property 5 ).

Series Lucas

They are the continuations where the relation of recurrence changed: it became

$U\_\; \{n+2\}\; =\; P\; \backslash \; cdot\; U\_n\; +\; Q\; \backslash \; cdot\; U\_\; \{n+1\}$

They are of two types according to whether initialization is of $u\_0\; =0$ and $u\_1=1$ or that it is $v\_0\; =2$ and $v\_1=P$. The continuation of Fibonacci is then a continuation U of Lucas of parameters P = 1 and Q = 1. The continuation of the numbers of Lucas is then a continuation v of Lucas of parameters $P\; =\; 1$ and $Q\; =\; 1$.

See also: Continuation of Lucas

Continuations of k-bonacci

These are continuations whose relation of recurrence is of order K . A term is the sum of the K terms which precede it

$u\_\; \{n+k\}\; \backslash ,\; =\; \backslash ,\; u\_n\; +\; u\_\; \{n+1\}\; +\; u\_\; \{n+2\}\; +\dots \; +u\_\; \{n+k\; -\; 1\}$

Among these continuations, one distinguishes the Suite from Tribonacci (recurrence of order 3) and the continuation of Tetranacci (recurrence of order 4). According to this new classification of continuations, the continuation of Fibonacci is a continuation of 2-bonacci.

If one modifies all at the same time (initialization, recurrence, order) one at the very general whole of the continuations at linear recurrence arrives.

See too

Related articles

• Leonardo Pisano

• Group of Fibonacci

External bonds

• Continuation of Fibonacci and golden section in the whole of Mandelbrot

• Continuation of Fibonacci in the dictionary of the numbers
• the blog of Gregory Pincus, which invents fibs , a form of Poésie using the continuation of Fibonacci
• Espiral de Durero there Las Meninas de Velázquez

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