# Bilangan Fibonacci

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Catatan: Penggunaan templat ini adalah tidak digalakkan.
Suatu ubinan dengan segi empat yang tepinya adalah bilangan Fibonaci berturut-turut pada panjangnya
Sebuah yupana (Quechua untuk "alat pengiraan") adalah sebuah kalkulator yang digunakan oleh Incas. Pengaji menganggapkan bahawa pengiraan adalah berasaskan bilangan Fibonacci untuk mengurangkan bilangan biji yang diperlukan tiap sawah.[1]
Sebuah Lingkaran Fibonacci dicipta dengan melukis lengkung menyambungkan sudut berlawan segi empat dalam ubinan Fibonacci; yang ini menggunakan segi empat-segi empat pada saiz 1, 1, 2, 3, 5, 8, 13, 21, and 34; see Golden spiral

Dalam matematik, bilangan Fibonacci adalah suatu langkah bilangan dinamakan selepas Leonardo of Pisa, digelar sebagai Fibonacci. Buku 1202 Liber Abaci Fibonacci memperkenalkan urutannya ke matematik Eropah Barat, walaupun urutannya telah terdahulu dijelaskan pada matematik India.[2][3]

Nombor urutan pertama adalah 0, nombor kedua adalah 1, dan setiap nombor subsequent bersamaan dengan jumlah dua nombor yang terdahulu pada urutannya sendiri. Dalam istilah matematik, ia ditakrifkan dengan recurrence relation yang berikut:

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

Iaitu, selepas dua nilai bermula, setiap nombow adalah jumlah dua nombor yang terdahulu. Bilangan Fibonacci pertama Templat:OEIS, juga ditandakan sebagai Fn, untuk n = 0, 1, 2, … ,20 adalah:[4][5]

 F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765

Setiap nombor ke-3 urutan adalah sama rata dan lebih umumnya, setiap nombor ke-k pada urutan adalah suatu perdaraban Fk.

Urutannya extended ke indeks negatif n memuaskan Fn = Fn−1 + Fn−2 untuk semua integer n, dan F−n = (−1)n+1Fn:

.., −8, 5, −3, 2, −1, 1, diikuti oleh urutan di atas.

## Origins

Builangan Fibonacci pertama kali muncul, di bawah nama mātrāmeru (gunung cadence), dalam karya Sanskrit grammarian Pingala (Chandah-shāstra, the Art of Prosody, 450 or 200 BC). Prosody adalah penting dalam upacara India silam oleh kerana suatu emfasis pada keaslian utterance. Ahli matematik Indian matematik Virahanka (abad ke-6 M) menunjukkan urutan Fibonacci berpunca pada analisis meter dengan silabel panjang dan pendek. Berikutnya itu, ahli falsafah Jain Hemachandra (sekitar 1150) mendirikan suatu teks diketahui benar pada ini. Suatu komen pada karya Virahanka oleh Gopāla pada abad ke-12 juga melawat semula masalah itu dalam sesetengah perincian.

Sanskrit vowel sounds can be long (L) or short (S), and Virahanka's analysis, which came to be known as mātrā-vṛtta, wishes to compute how many metres (mātrās) of a given overall length can be composed of these syllables. If the long syllable is twice as long as the short, the solutions are:

1 mora: S (1 corak)
2 morae: SS; L (2)
3 morae: SSS, SL; LS (3)
4 morae: SSSS, SSL, SLS; LSS, LL (5)
5 morae: SSSSS, SSSL, SSLS, SLSS, SLL; LSSS, LSL, LLS (8)
6 morae: SSSSSS, SSSSL, SSSLS, SSLSS, SLSSS, LSSSS, SSLL, SLSL, SLLS, LSSL, LSLS, LLSS, LLL (13)
7 morae: SSSSSSS, SSSSSL, SSSSLS, SSSLSS, SSLSSS, SLSSSS, LSSSSS, SSSLL, SSLSL, SLSSL, LSSSL, SSLLS, SLSLS, LSSLS, SLLSS, LSLSS, LLSSS, SLLL, LSLL, LLSL, LLLS (21)

A pattern of length n can be formed by adding S to a pattern of length n − 1, or L to a pattern of length n − 2; and the prosodicists showed that the number of patterns of length n is the sum of the two previous numbers in the sequence. Donald Knuth reviews this work in The Art of Computer Programming as equivalent formulations of the bin packing problem for items of lengths 1 and 2.

In the West, the sequence was first studied by Leonardo of Pisa, known as Fibonacci, in his Liber Abaci (1202)[6]. He considers the growth of an idealised (biologically unrealistic) rabbit population, assuming that:

• In the "zeroth" month, there is one pair of rabbits (additional pairs of rabbits = 0)
• In the first month, the first pair begets another pair (additional pairs of rabbits = 1)
• In the second month, both pairs of rabbits have another pair, and the first pair dies (additional pairs of rabbits = 1)
• In the third month, the second pair and the new two pairs have a total of three new pairs, and the older second pair dies. (additional pairs of rabbits = 2)

The laws of this are that each pair of rabbits has 2 pairs in its lifetime, and dies.

Let the population at month n be F(n). At this time, only rabbits who were alive at month n − 2 are fertile and produce offspring, so F(n − 2) pairs are added to the current population of F(n − 1). Thus the total is F(n) = F(n − 1) + F(n − 2).[7]

## Relation to the Golden Ratio

### Closed form expression

Like every sequence defined by linear recurrence, the Fibonacci numbers have a closed-form solution. It has become known as Binet's formula, even though it was already known by Abraham de Moivre:

$F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}={{\varphi^n-(-1/\varphi)^{n}} \over {\sqrt 5}}\, ,$ where $\varphi$ is the golden ratio
$\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,$ Templat:OEIS

(note, that $1-\varphi=-1/\varphi$, as can be seen from the defining equation below).

The Fibonacci recursion

$F(n+2)-F(n+1)-F(n)=0\,$

is similar to the defining equation of the golden ratio in the form

$x^2-x-1=0,\,$

which is also known as the generating polynomial of the recursion.

#### Proof by induction

Any root of the equation above satisfies $\begin{matrix}x^2=x+1,\end{matrix}\,$ and multiplying by $x^{n-1}\,$ shows:

$x^{n+1} = x^n + x^{n-1}\,$

By definition $\varphi$ is a root of the equation, and the other root is $1-\varphi=-1/\varphi\, .$. Therefore:

$\varphi^{n+1} = \varphi^n + \varphi^{n-1}\,$

and

$(1-\varphi)^{n+1} = (1-\varphi)^n + (1-\varphi)^{n-1}\, .$

Both $\varphi^{n}$ and $(1-\varphi)^{n}=(-1/\varphi)^{n}$ are geometric series (for n = 1, 2, 3, ...) that satisfy the Fibonacci recursion. The first series grows exponentially; the second exponentially tends to zero, with alternating signs. Because the Fibonacci recursion is linear, any linear combination of these two series will also satisfy the recursion. These linear combinations form a two-dimensional linear vector space; the original Fibonacci sequence can be found in this space.

Linear combinations of series $\varphi^{n}$ and $(1-\varphi)^{n}$, with coefficients a and b, can be defined by

$F_{a,b}(n) = a\varphi^n+b(1-\varphi)^n$ for any real $a,b\, .$

All thus-defined series satisfy the Fibonacci recursion

\begin{align} F_{a,b}(n+1) &= a\varphi^{n+1}+b(1-\varphi)^{n+1} \\ &=a(\varphi^{n}+\varphi^{n-1})+b((1-\varphi)^{n}+(1-\varphi)^{n-1}) \\ &=a{\varphi^{n}+b(1-\varphi)^{n}}+a{\varphi^{n-1}+b(1-\varphi)^{n-1}} \\ &=F_{a,b}(n)+F_{a,b}(n-1)\,. \end{align}

Requiring that $F_{a,b}(0)=0$ and $F_{a,b}(1)=1$ yields $a=1/\sqrt 5$ and $b=-1/\sqrt 5$, resulting in the formula of Binet we started with. It has been shown that this formula satisfies the Fibonacci recursion. Furthermore, an explicit check can be made:

$F_{a,b}(0)=\frac{1}{\sqrt 5}-\frac{1}{\sqrt 5}=0\,\!$

and

$F_{a,b}(1)=\frac{\varphi}{\sqrt 5}-\frac{(1-\varphi)}{\sqrt 5}=\frac{-1+2\varphi}{\sqrt 5}=\frac{-1+(1+\sqrt 5)}{\sqrt 5}=1,$

establishing the base cases of the induction, proving that

$F(n)={{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}$ for all $n\, .$

Therefore, for any two starting values, a combination $a,b$ can be found such that the function $F_{a,b}(n)\,$ is the exact closed formula for the series.

#### Computation by rounding

Since $\begin{matrix}|1-\varphi|^n/\sqrt 5 < 1/2\end{matrix}$ for all $n\geq 0$, the number $F(n)$ is the closest integer to $\varphi^n/\sqrt 5\, .$ Therefore it can be found by rounding, or in terms of the floor function:

$F(n)=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor.$

### Limit of consecutive quotients

Johannes Kepler observed that the ratio of consecutive Fibonacci numbers converges. He wrote that "as 5 is to 8 so is 8 to 13, practically, and as 8 is to 13, so is 13 to 21 almost”, and concluded that the limit approaches the golden ratio $\varphi$.[8]

$\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\varphi,$

This convergence does not depend on the starting values chosen, excluding 0, 0.

Proof:

It follows from the explicit formula that for any real $a \ne 0, \, b \ne 0 \,$

\begin{align} \lim_{n\to\infty}\frac{F_{a,b}(n+1)}{F_{a,b}(n)} &= \lim_{n\to\infty}\frac{a\varphi^{n+1}-b(1-\varphi)^{n+1}}{a\varphi^n-b(1-\varphi)^n} \\ &= \lim_{n\to\infty}\frac{a\varphi-b(1-\varphi)(\frac{1-\varphi}{\varphi})^n}{a-b(\frac{1-\varphi}{\varphi})^n} \\ &= \varphi \end{align}

because $\bigl|{\tfrac{1-\varphi}{\varphi}}\bigr| < 1$ and thus $\lim_{n\to\infty}\left(\tfrac{1-\varphi}{\varphi}\right)^n=0 .$

### Decomposition of powers of the golden ratio

Since the golden ratio satisfies the equation

$\varphi^2=\varphi+1,\,$

this expression can be used to decompose higher powers $\varphi^n$ as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of $\varphi$ and 1. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients, thus closing the loop:

$\varphi^n=F(n)\varphi+F(n-1).$

This expression is also true for $n \, <\, 1 \,$ if the Fibonacci sequence $F(n) \,$ is extended to negative integers using the Fibonacci rule $F(n) = F(n-1) + F(n-2) . \,$

## Matrix form

A 2-dimensional system of linear difference equations that describes the Fibonacci sequence is

${F_{k+2} \choose F_{k+1}} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} {F_{k+1} \choose F_{k}}$

or

$\vec F_{k+1} = A \vec F_{k}.\,$

The eigenvalues of the matrix A are $\varphi\,\!$ and $(1-\varphi)\,\!$, and the elements of the eigenvectors of A, ${\varphi \choose 1}$ and ${1 \choose -\varphi}$, are in the ratios $\varphi\,\!$ and $(1-\varphi\,\!).$

This matrix has a determinant of −1, and thus it is a 2×2 unimodular matrix. This property can be understood in terms of the continued fraction representation for the golden ratio:

$\varphi =1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\;\;\ddots\,}}} \;.$

The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for $\varphi\,\!$, and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1.

The matrix representation gives the following closed expression for the Fibonacci numbers:

$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}.$

Taking the determinant of both sides of this equation yields Cassini's identity

$(-1)^n = F_{n+1}F_{n-1} - F_n^2.\,$

Additionally, since $A^n A^m=A^{m+n}$ for any square matrix $A$, the following identities can be derived:

${F_n}^2 + {F_{n-1}}^2 = F_{2n-1},\,$
$F_{n+1}F_{m} + F_n F_{m-1} = F_{m+n}.\,$

For the first one of these, there is a related identity:

$(2F_{n-1}+F_n)F_n = (F_{n-1}+F_{n+1})F_n = F_{2n}.\,$

For another way to derive the $F_{2n+k}$ formulas see the "EWD note" by Dijkstra[9].

## Recognizing Fibonacci numbers

The question may arise whether a positive integer $z$ is a Fibonacci number. Since $F(n)$ is the closest integer to $\varphi^n/\sqrt{5}$, the most straightforward, brute-force test is the identity

$F\bigg(\bigg\lfloor\log_\varphi(\sqrt{5}z)+\frac{1}{2}\bigg\rfloor\bigg)=z,$

which is true if and only if $z$ is a Fibonacci number.

Alternatively, a positive integer $z$ is a Fibonacci number if and only if one of $5z^2+4$ or $5z^2-4$ is a perfect square.[10]

A slightly more sophisticated test uses the fact that the convergents of the continued fraction representation of $\varphi$ are ratios of successive Fibonacci numbers, that is the inequality

$\bigg|\varphi-\frac{p}{q}\bigg|<\frac{1}{q^2}$

(with coprime positive integers $p$, $q$) is true if and only if $p$ and $q$ are successive Fibonacci numbers. From this one derives the criterion that $z$ is a Fibonacci number if and only if the closed interval

$\bigg[\varphi z-\frac{1}{z},\varphi z+\frac{1}{z}\bigg]$

contains a positive integer.[11]

## Identities

Most identities involving Fibonacci numbers draw from combinatorial arguments. F(n) can be interpreted as the number of ways summing 1's and 2's to n − 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.

### First Identity

$F_{n+1} = F_{n} + F_{n-1}$
The nth Fibonacci number is the sum of the previous two Fibonacci numbers.

#### Proof

We must establish that the sequence of numbers defined by the combinatorial interpretation above satisfy the same recurrence relation as the Fibonacci numbers (and so are indeed identical to the Fibonacci numbers).

The set of F(n+1) ways of making ordered sums of 1's and 2's that sum to n may be divided into two non-overlapping sets. The first set contains those sums whose first summand is 1; the remainder sums to n−1, so there are F(n) sums in the first set. The second set contains those sums whose first summand is 2; the remainder sums to n−2, so there are F(n−1) sums in the second set. The first summand can only be 1 or 2, so these two sets exhaust the original set. Thus F(n+1) = F(n) + F(n−1).

### Second Identity

$\sum_{i=0}^n F_i = F_{n+2} - 1$
The sum of the first n Fibonacci numbers is the (n+2)nd Fibonacci number minus 1.

#### Proof

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 ≥ 0. Since there is only one sum of n + 1 that does not use any 2, namely 1 + … + 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 − 1. If the second summand is 2 but the first is 1, then there are F(n − 1) ways to complete the counting for n − 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 − 1) + … + F(0) gives the desired counting.

### Third Identity

This identity has slightly different forms for $F_k$, depending on whether k is odd or even.

$\sum_{i=0}^{n-1} F_{2i+1} = F_{2n}$
$\sum_{i=0}^{n} F_{2i} = F_{2n+1}-1$

[12]

The sum of the first n-1 Fibonacci numbers, $F_j$, such that j is odd is the (2n)th Fibonacci number.
The sum of the first n Fibonacci numbers, $F_j$, such that j is even is the (2n+1)th Fibonacci number minus 1.

#### Proofs

By induction for $F_{2n}$:

$F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}=F_{2n}$
$F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n}+F_{2n+1}$
$F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n+2}$

A basis case for this could be $F_1=F_2$.
By induction for $F_{2n+1}$:

$F_0+F_2+F_4+...+F_{2n-2}+F_{2n}=F_{2n+1}-1$
$F_0+F_2+F_4+...+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+1}+F_{2n+2}-1$
$F_0+F_2+F_4+...+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+3}-1$

A basis case for this could be $F_0=F_1-1$.

### Fourth Identity

$\sum_{i=0}^n iF_i = nF_{n+2} - F_{n+3} + 2$

#### Proof

This identity can be established in two stages. First, we count the number of ways summing 1s and 2s to −1, 0, …, or n + 1 such that at least one of the summands is 2.

By our second identity, there are F(n + 2) − 1 ways summing to n + 1; F(n + 1) − 1 ways summing to n; …; and, eventually, F(2) − 1 way summing to 1. As F(1) − 1 = F(0) = 0, we can add up all n + 1 sums and apply the second identity again to obtain

[F(n + 2) − 1] + [F(n + 1) − 1] + … + [F(2) − 1]
= [F(n + 2) − 1] + [F(n + 1) − 1] + … + [F(2) − 1] + [F(1) − 1] + F(0)
= F(n + 2) + [F(n + 1) + … + F(1) + F(0)] − (n + 2)
= F(n + 2) + [F(n + 3) − 1] − (n + 2)
= F(n + 2) + F(n + 3) − (n + 3).

On the other hand, we observe from the second identity that there are

• F(0) + F(1) + … + F(n − 1) + F(n) ways summing to n + 1;
• F(0) + F(1) + … + F(n − 1) ways summing to n;

……

• F(0) way summing to −1.

Adding up all n + 1 sums, we see that there are

• (n + 1) F(0) + n F(1) + … + F(n) ways summing to −1, 0, …, or n + 1.

Since the two methods of counting refer to the same number, we have

(n + 1) F(0) + n F(1) + … + F(n) = F(n + 2) + F(n + 3) − (n + 3)

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

### Fifth Identity

$\sum_{i=0}^n {F_i}^2 = F_{n} F_{n+1}$
The sum of the first n Fibonacci numbers squared is the product of the nth and (n+1)th Fibonacci numbers.

### Identity for doubling n

$F_{2n} = F_{n+1}^2 - F_{n-1}^2 = F_n(F_{n+1}+F_{n-1})$

[13]

### Another Identity

Another identity useful for calculating Fn for large values of n is

$F_{kn+c} = \sum_{i=0}^k {k\choose i} F_{c-i} F_n^i F_{n+1}^{k-i},$

[13]

from which other identities for specific values of k, n, and c can be derived below, including

$F_{2n+k} = F_k F_{n+1}^2 + 2 F_{k-1} F_{n+1} F_n + F_{k-2} F_n^2$

for all integers n and k. Dijkstra[9] points out that doubling identities of this type can be used to calculate Fn using O(log n) arithmetic operations. Notice that, with the definition of Fibonacci numbers with negative n given in the introduction, this formula reduces to the double n formula when k = 0.

(From practical standpoint it should be noticed that the calculation involves manipulation of numbers with length (number of digits) ${\rm \Theta}(n)\,$. Thus the actual performance depends mainly upon efficiency of the implemented long multiplication, and usually is ${\rm \Theta}(n \,\log n)$ or ${\rm \Theta}(n ^{\log_2 3})$.)

### Other identities

Other identities include relationships to the Lucas numbers, which have the same recursive properties but start with L0=2 and L1=1. These properties include F2n=FnLn.

There are also scaling identities, which take you from Fn and Fn+1 to a variety of things of the form Fan+b; for instance

$F_{3n} = 2F_n^3 + 3F_n F_{n+1} F_{n-1} = 5F_{n}^3 + 3 (-1)^n F_{n}$ by Cassini's identity.

$F_{3n+1} = F_{n+1}^3 + 3 F_{n+1}F_n^2 - F_n^3$

$F_{3n+2} = F_{n+1}^3 + 3 F_{n+1}^2F_n + F_n^3$

$F_{4n} = 4F_nF_{n+1}(F_{n+1}^2 + 2F_n^2) - 3F_n^2(F_n^2 + 2F_{n+1}^2)$

These can be found experimentally using lattice reduction, and are useful in setting up the special number field sieve to factorize a Fibonacci number. Such relations exist in a very general sense for numbers defined by recurrence relations, see the section on multiplication formulae under Perrin numbers for details.

## Power series

Fungsi generasi urutan Fibonacci adalah siri tenaga

$s(x)=\sum_{k=0}^{\infty} F_k x^k.$

Siri ini adalah mudah dan jawapan bentuk-tertutup menarik untuk $|x| < 1/\varphi$

$s(x)=\frac{x}{1-x-x^2}.$

Jawapan ini dapat dibukti dengan menggunakan kemunculan semula Fibonacci untuk melebarkan setiap koefisi dalam jumlah infinite mentakrifkan $s(x)$:

\begin{align} s(x) &= \sum_{k=0}^{\infty} F_k x^k \\ &= F_0 + F_1x + \sum_{k=2}^{\infty} \left( F_{k-1} + F_{k-2} \right) x^k \\ &= x + \sum_{k=2}^{\infty} F_{k-1} x^k + \sum_{k=2}^{\infty} F_{k-2} x^k \\ &= x + x\sum_{k=0}^{\infty} F_k x^k + x^2\sum_{k=0}^{\infty} F_k x^k \\ &= x + x s(x) + x^2 s(x) \end{align}

Menyelesaikan persamaan $s(x)=x+xs(x)+x^2s(x)$ for $s(x)$ menyebabkan jawapan bentuk tertutup.

Terutamanya, buku puzzle matematik menyatakan nilai curious $\frac{s(\frac{1}{10})}{10}=\frac{1}{89}$, atau lebih biasanya

$\sum_{n = 1}^{\infty}{\frac {F(n)}{10^{(k + 1)(n + 1)}}} = \frac {1}{10^{2k + 2} - 10^{k + 1} - 1}$

untuk semua integer $k >= 0$.

Secara bicara,

$\sum_{n=0}^\infty\,\frac{F_n}{k^{n}}\,=\,\frac{k}{k^{2}-k-1}.$

## Reciprocal sums

Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of theta functions. For example, we can write the sum of every odd-indexed reciprocal Fibonacci number as

$\sum_{k=0}^\infty \frac{1}{F_{2k+1}} = \frac{\sqrt{5}}{4}\vartheta_2^2 \left(0, \frac{3-\sqrt 5}{2}\right) ,$

and the sum of squared reciprocal Fibonacci numbers as

$\sum_{k=1}^\infty \frac{1}{F_k^2} = \frac{5}{24} \left(\vartheta_2^4\left(0, \frac{3-\sqrt 5}{2}\right) - \vartheta_4^4\left(0, \frac{3-\sqrt 5}{2}\right) + 1 \right).$

If we add 1 to each Fibonacci number in the first sum, there is also the closed form

$\sum_{k=0}^\infty \frac{1}{1+F_{2k+1}} = \frac{\sqrt{5}}{2},$

and there is a nice nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio,

$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2}.$

Results such as these make it plausible that a closed formula for the plain sum of reciprocal Fibonacci numbers could be found, but none is yet known. Despite that, the reciprocal Fibonacci constant

$\psi = \sum_{k=1}^{\infty} \frac{1}{F_k} = 3.359885666243 \dots$

has been proved irrational by Richard André-Jeannin.

## Primes and divisibility

Rencana utama: Fibonacci prime

A Fibonacci prime is a Fibonacci number that is prime Templat:OEIS. The first few are:

2, 3, 5, 13, 89, 233, 1597, 28657, 514229, …

Fibonacci primes with thousands of digits have been found, but it is not known whether there are infinitely many. They must all have a prime index, except F4 = 3. There are arbitrarily long runs of composite numbers and therefore also of composite Fibonacci numbers.

With the exceptions of 1, 8 and 144 (F0 = F1, F6 and F12) every Fibonacci number has a prime factor that is not a factor of any smaller Fibonacci number (Carmichael's theorem).[14]

No Fibonacci number greater than F6 = 8 is one greater or one less than a prime number.[15]

Any three consecutive Fibonacci numbers, taken two at a time, are relatively prime: that is,

gcd(Fn, Fn+1) = gcd(Fn, Fn+2) = 1.

More generally,

gcd(Fn, Fm) = Fgcd(n, m).[16][17]

### Odd divisors

If n is odd all the odd divisors of Fn are ≡ 1 (mod 4).[18][19]
This is equivalent to saying that for odd n all the odd prime factors of Fn are ≡ 1 (mod 4).

For example,

F1 = 1, F3 = 2, F5 = 5, F7 = 13, F9 = 34 = 2×17, F11 = 89, F13 = 233, F15 = 610 = 2×5×61

### Fibonacci and Legendre

There are some interesting formulas connecting the Fibonacci numbers and the Legendre symbol $\;\left(\tfrac{p}{5}\right).$

$\left(\frac{p}{5}\right) = \left \{ \begin{array}{cl} 0 & \textrm{if}\;p =5 \\ 1 &\textrm{if}\;p \equiv \pm1 \pmod 5 \\ -1 &\textrm{if}\;p \equiv \pm2 \pmod 5 \end{array} \right.$

If p is a prime number then[20][21] $F_{p} \equiv \left(\frac{p}{5}\right) \pmod p \;\;\mbox{ and }\;\;\; F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p.$

For example,

$(\tfrac{2}{5}) = -1, \,\, F_3 = 2, F_2=1,$
$(\tfrac{3}{5}) = -1, \,\, F_4 = 3,F_3=2,$
$(\tfrac{5}{5}) = \;\;\,0,\,\, F_5 = 5,$
$(\tfrac{7}{5}) = -1, \,\,F_8 = 21,\;\;F_7=13,$
$(\tfrac{11}{5}) = +1, F_{10} = 55, F_{11}=89.$

Also, if p ≠ 5 is an odd prime number then: [22]

$5F^2_{\left(p \pm 1 \right) / 2} \equiv \begin{cases} \frac{5\left(\frac{p}{5}\right)\pm 5}{2} \pmod p & \textrm{if}\;p \equiv 1 \pmod 4\\ \\ \frac{5\left(\frac{p}{5}\right)\mp 3}{2} \pmod p & \textrm{if}\;p \equiv 3 \pmod 4 \end{cases}$

Examples of all the cases:

$p=7 \equiv 3 \pmod 4, \;\;(\tfrac{7}{5}) = -1, \frac{5(\frac{7}{5})+3}{2} =-1\mbox{ and }\frac{5(\frac{7}{5})-3}{2}=-4.$
$F_3=2 \mbox{ and } F_4=3.$
$5F_3^2=20\equiv -1 \pmod {7}\;\;\mbox{ and }\;\;5F_4^2=45\equiv -4 \pmod {7}$
$p=11 \equiv 3 \pmod 4, \;\;(\tfrac{11}{5}) = +1, \frac{5(\frac{11}{5})+3}{2} =4\mbox{ and }\frac{5(\frac{11}{5})- 3}{2}=1.$
$F_5=5 \mbox{ and } F_6=8.$
$5F_5^2=125\equiv 4 \pmod {11} \;\;\mbox{ and }\;\;5F_6^2=320\equiv 1 \pmod {11}$
$p=13 \equiv 1 \pmod 4, \;\;(\tfrac{13}{5}) = -1, \frac{5(\frac{13}{5})-5}{2} =-5\mbox{ and }\frac{5(\frac{13}{5})+ 5}{2}=0.$
$F_6=8 \mbox{ and } F_7=13.$
$5F_6^2=320\equiv -5 \pmod {13} \;\;\mbox{ and }\;\;5F_7^2=845\equiv 0 \pmod {13}$
$p=29 \equiv 1 \pmod 4, \;\;(\tfrac{29}{5}) = +1, \frac{5(\frac{29}{5})-5}{2} =0\mbox{ and }\frac{5(\frac{29}{5})+5}{2}=5.$
$F_{14}=377 \mbox{ and } F_{15}=610.$
$5F_{14}^2=710645\equiv 0 \pmod {29} \;\;\mbox{ and }\;\;5F_{15}^2=1860500\equiv 5 \pmod {29}$

### Divisibility by 11

$\sum_{k=n}^{n+9} F_{k} = 11 F_{n+6}$

For example, let n = 1:

F1+F2+...+F10 = 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143 = 11×13
n = 2:
F2+F3+...+F11 = 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 = 231 = 11×21
n = 3:
F3+F4+...+F12 = 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144= 374 = 11×34

In fact, the identity is true for all integers n, not just positive ones:

n = 0:

F0+F1+...+F9 = 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88 = 11×8
n = −1:

F−1+F0+...+F8 = 1 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 = 55 = 11×5
n = −2:

F−2+F−1+F0+...+F7 = −1 + 1 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 = 33 = 11×3

## Right triangles

Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple. The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle.

The first triangle in this series has sides of length 5, 4, and 3. Skipping 8, the next triangle has sides of length 13, 12 (5 + 4 + 3), and 5 (8 − 3). Skipping 21, the next triangle has sides of length 34, 30 (13 + 12 + 5), and 16 (21 − 5). This series continues indefinitely. The triangle sides a, b, c can be calculated directly:

$\displaystyle a_n = F_{2n-1}$
$\displaystyle b_n = 2 F_n F_{n-1}$
$\displaystyle c_n = {F_n}^2 - {F_{n-1}}^2$

These formulas satisfy $a_n ^2 = b_n ^2 + c_n ^2$ for all n, but they only represent triangle sides when $n > 2$.

Any four consecutive Fibonacci numbers Fn, Fn+1, Fn+2 and Fn+3 can also be used to generate a Pythagorean triple in a different way:

$a = F_n F_{n+3} \, ; \, b = 2 F_{n+1} F_{n+2} \, ; \, c = F_{n+1}^2 + F_{n+2}^2 \, ; \, a^2 + b^2 = c^2 \,.$

Example 1: let the Fibonacci numbers be 1, 2, 3 and 5. Then:

$\displaystyle a = 1 \times 5 = 5$
$\displaystyle b = 2 \times 2 \times 3 = 12$
$\displaystyle c = 2^2 + 3^2 = 13 \,$
$\displaystyle 5^2 + 12^2 = 13^2 \,.$

Example 2: let the Fibonacci numbers be 8, 13, 21 and 34. Then:

$\displaystyle a = 8 \times 34 = 272$
$\displaystyle b = 2 \times 13 \times 21 = 546$
$\displaystyle c = 13^2 + 21^2 = 610 \,$
$\displaystyle 272^2 + 546^2 = 610^2 \,.$

## Magnitude of Fibonacci numbers

Since $F_n$ is asymptotic to $\varphi^n/\sqrt5$, the number of digits in the base b representation of $F_n\,$ is asymptotic to $n\,\log_b\varphi$.

In base 10, for every integer greater than 1 there are 4 or 5 Fibonacci numbers with that number of digits, in most cases 5.

## Applications

The Fibonacci numbers are important in the run-time analysis of Euclid's algorithm to determine the greatest common divisor of two integers: the worst case input for this algorithm is a pair of consecutive Fibonacci numbers.

Yuri 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 the sums of "shallow" diagonals in Pascal's triangle and Lozanić's triangle (see "Binomial coefficient"). (They occur more obviously in Hosoya's triangle).

Every positive integer can be written in a unique way as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. This is known as Zeckendorf's theorem, and a sum of Fibonacci numbers that satisfies these conditions is called a Zeckendorf representation.

The Fibonacci numbers and principle is also used in the financial markets. It is used in trading algorithms, applications and strategies. Some typical forms include: the Fibonacci fan, Fibonacci Arc, Fibonacci Retracement and the Fibonacci Time Extension.

Fibonacci numbers are used by some pseudorandom number generators.

Fibonacci numbers are used in a polyphase version of the merge sort algorithm in which an unsorted list is divided into two lists whose lengths correspond to sequential Fibonacci numbers - by dividing the list so that the two parts have lengths in the approximate proportion φ. A tape-drive implementation of the polyphase merge sort was described in The Art of Computer Programming.

Fibonacci numbers arise in the analysis of the Fibonacci heap data structure.

A one-dimensional optimization method, called the Fibonacci search technique, uses Fibonacci numbers.[23]

The Fibonacci number series is used for optional lossy compression in the IFF 8SVX audio file format used on Amiga computers. The number series compands the original audio wave similar to logarithmic methods e.g. µ-law.[24][25]

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. It is commonly thought that the first movement of Béla Bartók's Music for Strings, Percussion, and Celesta was structured using Fibonacci numbers.

Since the conversion factor 1.609344 for miles to kilometers is close to the golden ratio (denoted φ), the decomposition of distance in miles into a sum of Fibonacci numbers becomes nearly the kilometer sum when the Fibonacci numbers are replaced by their successors. This method amounts to a radix 2 number register in golden ratio base φ being shifted. To convert from kilometers to miles, shift the register down the Fibonacci sequence instead.[26][27][28]

## Fibonacci numbers in nature

Sunflower head displaying florets in spirals of 34 and 55 around the outside

Fibonacci sequences appear in biological settings,[29] in two consecutive Fibonacci numbers, such as branching in trees, arrangement of leaves on a stem, the fruitlets of a pineapple,[30] the flowering of artichoke, an uncurling fern and the arrangement of a pine cone.[31] In addition, numerous poorly substantiated claims of Fibonacci numbers or golden sections in nature are found in popular sources, e.g. relating to the breeding of rabbits, the spirals of shells, and the curve of waves[perlu rujukan]. The Fibonacci numbers are also found in the family tree of honeybees. [32]

Przemyslaw Prusinkiewicz advanced the idea that real instances can be in part understood as the expression of certain algebraic constraints on free groups, specifically as certain Lindenmayer grammars.[33]

A model for the pattern of florets in the head of a sunflower was proposed by H. Vogel in 1979.[34] This has the form

$\theta = \frac{2\pi}{\phi^2} n$, $r = c \sqrt{n}$

where n is the index number of the floret and c is a constant scaling factor; the florets thus lie on Fermat's spiral. The divergence angle, approximately 137.51°, is the golden angle, dividing the circle in the golden ratio. Because this ratio is irrational, no floret has a neighbor at exactly the same angle from the center, so the florets pack efficiently. Because the rational approximations to the golden ratio are of the form F(j):F(j+1), the nearest neighbors of floret number n are those at n±F(j) for some index j which depends on r, the distance from the center. It is often said that sunflowers and similar arrangements have 55 spirals in one direction and 89 in the other (or some other pair of adjacent Fibonacci numbers), but this is true only of one range of radii, typically the outermost and thus most conspicuous.[35]

## Generalizations

The Fibonacci sequence has been generalized in many ways. These include:

• Generalizing the index to negative integers to produce the Negafibonacci numbers.
• Generalizing the index to real numbers using a modification of Binet's formula. [37]
• Starting with other integers. Lucas numbers have L1 = 1, L2 = 3, and Ln = Ln−1 + Ln−2. Primefree sequences use the Fibonacci recursion with other starting points in order to generate sequences in which all numbers are composite.
• Letting a number be a linear function (other than the sum) of the 2 preceding numbers. The Pell numbers have Pn = 2Pn – 1 + Pn – 2.
• Not adding the immediately preceding numbers. The Padovan sequence and Perrin numbers have P(n) = P(n – 2) + P(n – 3).
• Generating the next number by adding 3 numbers (tribonacci numbers), 4 numbers (tetranacci numbers), or more.
• Adding other objects than integers, for example functions or strings -- one essential example is Fibonacci polynomials.

## Numbers properties

### Periodicity mod n: Pisano periods

It is easily seen that if the members of the Fibonacci sequence are taken mod n, the resulting sequence must be periodic with period at most $n^2$. The lengths of the periods for various n form the so-called Pisano periods Templat:OEIS. Determining the Pisano periods in general is an open problem,[perlu rujukan] although for any particular n it can be solved as an instance of cycle detection.

## The bee ancestry code

Fibonacci numbers also appear in the description of the reproduction of a population of idealized bees, according to the following rules:

• If an egg is laid by an unmated female, it hatches a male.
• If, however, an egg was fertilized by a male, it hatches a female.

Thus, a male bee will always have one parent, and a female bee will have two.

If one traces the ancestry of any male bee (1 bee), he has 1 female parent (1 bee). This female had 2 parents, a male and a female (2 bees). The female had two parents, a male and a female, and the male had one female (3 bees). Those two females each had two parents, and the male had one (5 bees). This sequence of numbers of parents is the Fibonacci sequence.[38]

This is an idealization that does not describe actual bee ancestries. In reality, some ancestors of a particular bee will always be sisters or brothers, thus breaking the lineage of distinct parents.

## Miscellaneous

In 1963, John H. E. Cohn proved that the only squares among the Fibonacci numbers are 0, 1, and 144.[39]

In 1990, Jean-claude Perez published strong links between fractals world and Fibonacci numbers sensitivity [40][41]

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2. Parmanand Singh. "Acharya Hemachandra and the (so called) Fibonacci Numbers". Math. Ed. Siwan, 20(1):28-30, 1986. ISSN 0047-6269]
3. Parmanand Singh,"The So-called Fibonacci numbers in ancient and medieval India." Historia Mathematica 12(3), 229–44, 1985.
4. Mengikut konvensyen moden, urutannya bermula dengan F0=0. Liber Abaci memulakan urutan dengan F1 = 1, meninggalkan permulaan 0, dan urutannya masih ditulis secara ini oleh sesetengah.
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7. Knott, Ron. "Fibonacci's Rabbits". University of Surrey School of Electronics and Physical Sciences.
8. Kepler, Johannes (1966). A New Year Gift: On Hexagonal Snow. Oxford University Press. m/s. 92. ISBN 0198581203.  Strena seu de Nive Sexangula (1611)
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11. M. Möbius, Wie erkennt man eine Fibonacci Zahl?, Math. Semesterber. (1998) 45; 243–246
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17. Su, Francis E., et al. "Fibonacci GCD's, please.", Mudd Math Fun Facts.
18. Lemmermeyer, ex. 2.27 p. 73
19. The website [2] has the first 300 Fibonacci numbers factored into primes.
20. Paulo Ribenboim (1996), The New Book of Prime Number Records, New York: Springer, ISBN 0-387-94457-5, p. 64
21. Franz Lemmermeyer (2000), Reciprocity Laws, New York: Springer, ISBN 3-540-66957-4, ex 2.25-2.28, pp. 73-74
22. Lemmermeyer, ex. 2.38, pp. 73-74
23. M. Avriel and D.J. Wilde (1966). "Optimality of the Symmetric Fibonacci Search Technique". Fibonacci Quarterly (3): 265–269.
24. Amiga ROM Kernel Reference Manual, Addison-Wesley 1991
25. IFF - MultimediaWiki
26. An Application of the Fibonacci Number Representation
27. A Practical Use of the Sequence
28. Zeckendorf representation
29. S. Douady and Y. Couder (1996). "Phyllotaxis as a Dynamical Self Organizing Process" (PDF). Journal of Theoretical Biology 178 (178): 255–274. doi:10.1006/jtbi.1996.0026.
30. Jones, Judy; William Wilson (2006). "Science". An Incomplete Education. Ballantine Books. m/s. 544. ISBN 978-0-7394-7582-9.
31. A. Brousseau (1969). "Fibonacci Statistics in Conifers". Fibonacci Quarterly (7): 525–532.
32. Computer Science for Fun - cs4fn: Marks for the da Vinci Code: B
33. Prusinkiewicz, Przemyslaw; James Hanan (1989). Lindenmayer Systems, Fractals, and Plants (Lecture Notes in Biomathematics). Springer-Verlag. ISBN 0-387-97092-4.
34. Vogel, H (1979), "A better way to construct the sunflower head", Mathematical Biosciences 44 (44): 179–189, doi:10.1016/0025-5564(79)90080-4
35. . Springer-Verlag. m/s. 101-107. ISBN 978-0387972978.
36. J.C. Perez (1991), "Chaos DNA and Neuro-computers: A Golden Link", in Speculations in Science and Technology vol. 14 no. 4, ISSN 0155-7785
37. Pravin Chandra and Eric W. Weisstein, Fibonacci Number di MathWorld.
38. The Fibonacci Numbers and the Ancestry of Bees
39. J H E Cohn. "Square Fibonacci Numbers Etc", pp. 109-113.