Revision 3684616 of "Nombor Fibonacci" on mswiki

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[[Image:FibonacciBlocks.svg|thumb|180px|right|Suatu ubinan dengan segi empat yang tepinya adalah nombor Fibonaci berturut-turut pada panjangnya]]
[[Image:Yupana 1.png|thumb|180px|right|Sebuah '''yupana''' ([[Quechua]] untuk "alat pengiraan") adalah sebuah kalkulator yang digunakan oleh [[Incas]]. Pengaji menganggapkan bahawa pengiraan adalah berasaskan nombor Fibonacci untuk mengurangkan bilangan biji yang diperlukan tiap sawah.<ref>http://www.quipus.it/english/Andean%20Calculators.pdf</ref>]]
[[Image:Fibonacci spiral 34.svg|right|thumb|180px|Sebuah [[Nombor Fibonacci|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]], '''nombor Fibonacci''' adalah suatu [[langkah]] nombor dinamakan sempena [[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]].<ref>Parmanand Singh. "Acharya Hemachandra and the (so called) Fibonacci Numbers". Math. Ed. Siwan, 20(1):28-30, 1986. ISSN 0047-6269]</ref><ref>Parmanand Singh,"The So-called Fibonacci numbers in ancient and medieval India." Historia Mathematica 12(3), 229–44, 1985.</ref>

Nombor urutan pertama adalah 0, nombor kedua adalah 1, dan setiap nombor seterusnya bersamaan dengan jumlah dua nombor yang terdahulu pada urutannya sendiri. Dalam istilah matematik, ia ditakrifkan dengan [[hubungan jadi semula]] yang berikut:

:<math>
  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}
 </math>

Iaitu, selepas dua nilai bermula, setiap nombor adalah jumlah dua nombor yang terdahulu. Nombor Fibonacci pertama  {{OEIS|id=A000045}}, juga ditandakan sebagai ''F<sub>n</sub>'', untuk ''n''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2, … ,20 adalah:<ref> Mengikut konvensyen moden, urutannya bermula dengan ''F''<sub>0</sub>=0. ''Liber Abaci'' memulakan urutan dengan ''F''<sub>1</sub> = 1, meninggalkan permulaan 0, dan urutannya masih ditulis secara ini oleh sesetengah.</ref><ref>The website [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibtable.html] has the first 300 F<sub>''n''</sub> factored into primes and links to more extensive tables.</ref>
:{| class="wikitable"
|-
| ''F''<sub>0</sub>
| ''F''<sub>1</sub>
| ''F''<sub>2</sub>
| ''F''<sub>3</sub>
| ''F''<sub>4</sub>
| ''F''<sub>5</sub>
| ''F''<sub>6</sub>
| ''F''<sub>7</sub>
| ''F''<sub>8</sub>
| ''F''<sub>9</sub>
| ''F''<sub>10</sub>
| ''F''<sub>11</sub>
| ''F''<sub>12</sub>
| ''F''<sub>13</sub>
| ''F''<sub>14</sub>
| ''F''<sub>15</sub>
| ''F''<sub>16</sub>
| ''F''<sub>17</sub>
| ''F''<sub>18</sub>
| ''F''<sub>19</sub>
| ''F''<sub>20</sub>
|-
| 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 ''F<sub>k</sub>''.

Urutannya extended ke indeks negatif ''n'' memuaskan ''F<sub>n</sub>'' = ''F''<sub>''n''−1</sub>&nbsp;+&nbsp;''F''<sub>''n''−2</sub> untuk ''semua'' integer ''n'', dan ''F<sub>−n</sub>'' = (−1)<sup>''n''+1</sup>''F''<sub>''n''</sub>:

.., &minus;8, 5, &minus;3, 2, &minus;1, 1, diikuti oleh urutan di atas.

== Asal Usul == 
Nombor Fibonacci pertama kali muncul, di bawah nama ''mātrāmeru'' (gunung [[irama]]), dalam karya [[ahli tatabahasa]] [[Pingala]] (''Chandah-shāstra'', Seni Prosodi, [[450 BC|450]] or [[200 BC]]).  [[Prosody (linguistics)|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 Veda|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 [[Gopala (ahli matematik)|Gopāla]] pada abad ke-12 juga melawat semula masalah itu dalam sesetengah perincian.

Bunyi vokal Sanskrit boleh menjadi panjang (L) atau pendek (S), dan analisis Virahanka, yang kemudian dikenali sebagai ''mātrā-vṛtta'', ingin mengira berapa meter (''mātrā'') bagi panjang keseluruhan yang boleh terdiri daripada silabel  ini. Jika silabel  panjang adalah dua kali lebih panjang berbanding yang pendek, penyelesaian ialah:
: 1 [[mora (linguistik)|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)

Satu corak panjang ''n'' boleh dibentuk dengan menambah S kepada corak panjang ''n''&nbsp;−&nbsp;1, atau L kepada corak panjang ''n''&nbsp;−&nbsp;2; dan pakar prosodi menunjukkan bahawa bilangan corak panjang ''n'' adalah jumlah dua nombor sebelumnya dalam urutan.  [[Donald Knuth]] menyemak kerja ini dalam ''[[The Art of Computer Programming]]'' <!-- see (Vol.&nbsp;1, &sect;1.2.8: Fibonacci Numbers)--> sebagai rumusan bersamaan [[masalah bin packing]] item dengan panjang 1 dan 2.

Di Barat, urutan itu mula-mula dikaji oleh Leonardo dari Pisa, dikenali sebagai [[Fibonacci]], di dalam bukunya [[Liber Abaci]] ([[1202]])<ref>{{cite book | title = Fibonacci's Liber Abaci | author = Sigler, Laurence E. (trans.) | publisher = Springer-Verlag | year = 2002 | id = ISBN 0-387-95419-8}} Chapter II.12, pp. 404–405.</ref>.  Dia menganggap pertumbuhan unggul  populasi arnab (secara biologinya tidak realistik), dengan anggapan bahawa:
* Dalam bulan "sifar", ada sepasang arnab (pasangan tambahan arnab&nbsp;=&nbsp;0)
* Dalam bulan pertama, pasangan pertama beranak sepasang lagi (pasangan tambahan arnab&nbsp;=&nbsp;1)
* Dalam bulan kedua, kedua-dua pasang arnab mempunyai sepasang lagi, dan pasangan yang pertama mati (pasangan tambahan arnab&nbsp;=&nbsp;1)
* Dalam bulan ketiga, pasangan kedua dan kedua-dua pasangan baru mempunyai sejumlah tiga pasangan baru, dan pasangan kedua tua mati. (pasangan tambahan arnab&nbsp;=&nbsp;2)

Hukum ini adalah bahawa setiap pasangan arnab mempunyai 2 pasang dalam hidupnya, dan mati.

Biarkan populasi pada bulan ''n'' menjadi ''F''(''n''). Pada masa ini, hanya arnab yang masih hidup pada bulan ''n''&nbsp;−&nbsp;2 adalah subur dan melahirkan anak, jadi ''F''(''n''&nbsp;−&nbsp;2) pasangan ditambah kepada populasi semasa ''F''(''n''&nbsp;−&nbsp;1). Oleh itu, jumlah ''F''(''n'')&nbsp;=&nbsp;''F''(''n''&nbsp;−&nbsp;1)&nbsp;+&nbsp;''F''(''n''&nbsp;−&nbsp;2).<ref>{{cite web
  | last = Knott
  | first = Ron
  | title = Fibonacci's Rabbits
  | url=http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibnat.html#Rabbits
  | publisher =[[University of Surrey]] School of Electronics and Physical Sciences}}</ref>

== Kaitannya dengan [[Nisbah Emas]] ==
===Closed form expression===
Like every sequence defined by linear [[Recurrence relation|recurrence]], the Fibonacci numbers have a  [[closed-form expression|closed-form solution]]. It has become known as [[Jacques Philippe Marie Binet|Binet]]'s formula, even though it was already known by [[Abraham de Moivre]]:
:<math>F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}={{\varphi^n-(-1/\varphi)^{n}} \over {\sqrt 5}}\, ,</math> where <math>\varphi</math> is the [[golden ratio]] 
:<math>\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\,39887\dots\,</math>  {{OEIS|id=A001622}}
(note, that <math>1-\varphi=-1/\varphi</math>, as can be seen from the defining equation below).

The Fibonacci recursion

:<math>F(n+2)-F(n+1)-F(n)=0\,</math>

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

:<math>x^2-x-1=0,\,</math>

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

====Proof by [[Mathematical induction|induction]]====
Any root of the equation above satisfies <math>\begin{matrix}x^2=x+1,\end{matrix}\,</math> and multiplying by <math>x^{n-1}\,</math> shows:
:<math>x^{n+1} = x^n + x^{n-1}\,</math>

By definition <math>\varphi</math> is a root of the equation, and the other root is <math>1-\varphi=-1/\varphi\, .</math>. Therefore:
:<math>\varphi^{n+1}  = \varphi^n + \varphi^{n-1}\, </math>

and
:<math>(1-\varphi)^{n+1} = (1-\varphi)^n + (1-\varphi)^{n-1}\, .</math>

Both <math>\varphi^{n}</math> and <math>(1-\varphi)^{n}=(-1/\varphi)^{n}</math>
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 <math>\varphi^{n}</math> and <math>(1-\varphi)^{n}</math>, with coefficients ''a'' and ''b'', can be defined by
:<math>F_{a,b}(n) = a\varphi^n+b(1-\varphi)^n</math> for any real <math>a,b\, .</math>

All thus-defined series satisfy the Fibonacci recursion
:<math>\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}</math>
Requiring that <math>F_{a,b}(0)=0</math> and <math>F_{a,b}(1)=1</math> yields <math>a=1/\sqrt 5</math> and <math>b=-1/\sqrt 5</math>, 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:
:<math>F_{a,b}(0)=\frac{1}{\sqrt 5}-\frac{1}{\sqrt 5}=0\,\!</math>

and
:<math>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,</math>

establishing the base cases of the induction, proving that
:<math>F(n)={{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}</math> for all <math> n\, .</math>

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

====Pengiraan melalui pembundaran====
Memandangkan <math>\begin{matrix}|1-\varphi|^n/\sqrt 5 < 1/2\end{matrix}</math> bagi semua <math>n\geq 0</math>, nombor <math>F(n)</math> adalah integer yang paling hampir dengan <math>\varphi^n/\sqrt 5\, .</math> Oleh itu, ia boleh didapati dengan [[Pembundaran#Pembundaran dalam pengiraan tepat|pembundaran]], atau dari segi [[fungsi lantai]]:
:<math>F(n)=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor.</math>

===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 <math>\varphi</math>.<ref>{{cite book | last=Kepler | first=Johannes | title=A New Year Gift: On Hexagonal Snow | year=1966 | isbn=0198581203 | publisher=Oxford University Press | pages=92}} Strena seu de Nive Sexangula (1611)</ref>

:<math>\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\varphi,</math>  
This convergence does not depend on the starting values chosen, excluding 0, 0.

'''Proof''':

It follows from the explicit formula that for any real <math>a \ne 0, \, b \ne 0 \,</math>
:<math>\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}</math>
because <math>\bigl|{\tfrac{1-\varphi}{\varphi}}\bigr| < 1</math> and thus <math>\lim_{n\to\infty}\left(\tfrac{1-\varphi}{\varphi}\right)^n=0 .</math>

===Decomposition of powers of the golden ratio===
Since the golden ratio satisfies the equation 
:<math>\varphi^2=\varphi+1,\,</math>
this expression can be used to decompose higher powers <math>\varphi^n</math> as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of <math>\varphi</math> and 1. The resulting [[recurrence relation]]ships yield Fibonacci numbers as the linear coefficients, thus closing the loop:
:<math>\varphi^n=F(n)\varphi+F(n-1).</math>
This expression is also true for <math>n \, <\, 1 \, </math> if the Fibonacci sequence <math>F(n) \,</math> is  [[Generalizations_of_Fibonacci_numbers#Extension_to_negative_integers|extended to negative integers]] using the Fibonacci rule <math>F(n) = F(n-1) + F(n-2) . \, </math>

==Matrix form==

A 2-dimensional system of linear [[difference equations]] that describes the Fibonacci sequence is
:<math>{F_{k+2} \choose F_{k+1}} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} {F_{k+1} \choose F_{k}}</math>

or
:<math>\vec F_{k+1} = A \vec F_{k}.\,</math>

The [[eigenvalue]]s of the matrix A are <math>\varphi\,\!</math> and <math>(1-\varphi)\,\!</math>, and the elements of the [[eigenvector]]s of A, <math>{\varphi \choose 1}</math> and <math>{1 \choose -\varphi}</math>, are in the ratios <math>\varphi\,\!</math> and <math>(1-\varphi\,\!).</math>

This matrix has a [[determinant]] of &minus;1, and thus it is a 2&times;2 [[unimodular matrix]].  This property can be understood in terms of the [[continued fraction]] representation for the golden ratio: 
:<math>\varphi
=1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\;\;\ddots\,}}} \;. </math> 
The Fibonacci numbers occur as the ratio of successive  convergents of the continued fraction for <math>\varphi\,\!</math>, and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or &minus;1.

The matrix representation gives the following [[closed expression]] for the Fibonacci numbers:
:<math>\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n =
       \begin{pmatrix} F_{n+1} & F_n \\
                       F_n     & F_{n-1} \end{pmatrix}.
</math>

Taking the determinant of both sides of this equation yields [[Cassini's identity]]
:<math>(-1)^n = F_{n+1}F_{n-1} - F_n^2.\,</math>

Additionally, since <math> A^n A^m=A^{m+n}</math> for any square matrix <math>A</math>, the following identities can be derived:
:<math>{F_n}^2 + {F_{n-1}}^2 = F_{2n-1},\,</math>
:<math>F_{n+1}F_{m} + F_n F_{m-1} = F_{m+n}.\, </math>

For the first one of these, there is a related identity:
:<math>(2F_{n-1}+F_n)F_n = (F_{n-1}+F_{n+1})F_n = F_{2n}.\,</math>
For another way to derive the <math>F_{2n+k}</math> formulas see the "EWD note" by [[Dijkstra]]<ref name="dijkstra78">E. W. Dijkstra (1978). ''In honour of Fibonacci.'' [http://www.cs.utexas.edu/users/EWD/ewd06xx/EWD654.PDF Report EWD654]</ref>.

==Recognizing Fibonacci numbers==

The question may arise whether a positive integer <math>z</math> is a Fibonacci number. Since <math>F(n)</math> is the closest integer to <math>\varphi^n/\sqrt{5}</math>, the most straightforward, brute-force test is the identity
:<math>F\bigg(\bigg\lfloor\log_\varphi(\sqrt{5}z)+\frac{1}{2}\bigg\rfloor\bigg)=z,</math>
which is true [[if and only if]] <math>z</math> is a Fibonacci number.

Alternatively, a positive integer <math>z</math> is a Fibonacci number if and only if one of <math>5z^2+4</math> or <math>5z^2-4</math> is a [[perfect square]].<ref>{{cite book | last=Posamentier | first=Alfred | coauthors = Lehmann, Ingmar| title=The (Fabulous) FIBONACCI Numbers | year=2007 | isbn=978-1-59102-475-0 | publisher=Prometheus Books | pages=305}}</ref> 

A slightly more sophisticated test uses the fact that the [[convergent (continued fraction)|convergent]]s of the [[continued fraction]] representation of <math>\varphi</math> are ratios of successive Fibonacci numbers, that is the inequality
:<math>\bigg|\varphi-\frac{p}{q}\bigg|<\frac{1}{q^2}</math>
(with [[coprime]] positive integers <math>p</math>, <math>q</math>) is true if and only if <math>p</math> and <math>q</math> are successive Fibonacci numbers. From this one derives the criterion that <math>z</math> is a Fibonacci number if and only if the [[closed interval]]
:<math>\bigg[\varphi z-\frac{1}{z},\varphi z+\frac{1}{z}\bigg]</math>
contains a positive integer.<ref>M.&nbsp;Möbius, ''Wie erkennt man eine Fibonacci Zahl?'', Math. Semesterber. (1998) 45; 243–246</ref>

==Pengenalan==

Kebanyakan pengenalan melibatkan nombor Fibonacci menarik dari [[bukti kombinatorik|hujah kombinatorik]].
''F''(''n'') boleh ditafsirkan sebagai bilangan cara menjumlahkan 1 dan 2 kepada ''n'' &minus; 1, dengan kelaziman yang ''F''(0) = 0, bermakna tiada jumlah akan menambah sehingga &minus;1, dan ''F''(1) = 1, bermaksud jumlah kosong akan "bertambah" untuk 0.
Ini tertiba bagi perihal penghasil tambah.
Sebagai contoh, 1 + 2 and 2 + 1 adalah dianggap dua jumlah yang berbeza dan dikira dua kali.

=== Pengenalan Pertama ===

:<big><math>F_{n+1} = F_{n} + F_{n-1}</math></big>

:''Nombor Fibonacci ke-n adalah jumlah dua nombor Fibonacci sebelumnya.''

==== Pembuktian====
Kita mesti membuktikan bahawa urutan nombor yang ditakrifkan oleh tafsiran kombinatorik di atas memenuhi hubungan jadi semula yang sama dengan nombor Fibonacci (dan jadi sememangnya sama dengan nombor Fibonacci).

Set bagi cara ''F''(''n''+1) untuk membuat jumlah bertertib 1 dan 2 yang berjumlah ke ''n'' boleh dibahagikan kepada dua set tak bertindih. Set pertama yang mengandungi jumlah penghasil tambah pertamanya 1; jumlah baki kepada ''n''&minus;1, maka ''F''(''n'') menjumlah pada set pertama. Set kedua yang mengandungi jumlah penghasil tambah pertamanya 2; jumlah baki kepada ''n''&minus;2, maka ''F''(''n''&minus;1) menjumlah pada set kedua. Penghasil tambah pertama hanya boleh jadi 1 atau 2, supaya kedua-dua set menghabiskan set asal. Maka ''F''(''n''+1) = ''F''(''n'') + ''F''(''n''&minus;1).

=== Pengenalan Kedua ===

:<math>\sum_{i=0}^n F_i = F_{n+2} - 1</math>

:''Jumlah bagi n pertama nombor Fibonacci adalah nombor Fibonacci ke-(n+2) tolak 1.''

==== Pembuktian ====

Kami mengira bilangan cara menjumlahkan 1 dan 2 sehingga ''n'' + 1 supaya sekurang-kurangnya salah satu daripada penghasil tambah adalah 2.

Seperti sebelum ini, terdapat ''F''(''n'' + 2) cara menjumlahkan 1 dan 2 menjadi ''n'' + 1 apabila ''n'' ≥ 0.
Oleh kerana hanya ada satu jumlah ''n'' + 1 yang tidak menggunakan 2, iaitu 1 + … + 1 (syarat ''n'' + 1),kita tolak 1 dari ''F''(''n'' + 2).

Setara, kita boleh mempertimbangkan sebutan pertama bagi 2 sebagai penghasil tambah.
Jika, dalam jumlah, penghasil tambah yang pertama adalah 2, maka ada ''F''(''n'') cara untuk melengkapkan pengiraan bagi ''n'' &minus; 1.
Jika penghasil tambah yang kedua ialah 2 tetapi yang pertama adalah 1, maka terdapat ''F''(''n'' &minus; 1) cara untuk melengkapkan pengiraan bagi ''n'' &minus; 2.
Teruskan dengan cara ini.
Kemudiannya kita mempertimbangkan penghasil tambah ke-(''n'' + 1).
Jika ia adalah 2 tetapi semua penghasil tambah''n'' sebelumnya adalah 1, maka terdapat ''F''(0) cara untuk melengkapkan pengiraan bagi 0.
Jika suatu jumlah mengandungi 2 sebagai penghasil tambah, sebutan pertama bagi penghasil tambah itu mesti berlaku di antara posisi yang pertama dan yang ke-(''n'' + 1).
Maka ''F''(''n'') + ''F''(''n'' &minus; 1) + … + ''F''(0) memberikan pengiraan yang dikehendaki.

=== Pengenalan Ketiga ===

Identiti ini mempunyai bentuk yang sedikit berbeza untuk <math>F_k</math>, bergantung kepada sama ada k adalah ganjil atau genap.
:<math>\sum_{i=0}^{n-1} F_{2i+1} = F_{2n}</math>
:<math>\sum_{i=0}^{n} F_{2i} = F_{2n+1}-1</math>

<ref>{{cite book | title = Fibonacci Numbers |last = Vorobiev |first = Nikolaĭ Nikolaevich |coauthors = Mircea Martin | publisher = Birkhäuser | year = 2002 | id = ISBN 3-7643-6135-2 |chapter=Chapter 1 |pages = pp. 5–6}}</ref>

:''Jumlah bagi nombor Fibonacci n-1 pertama, <math>F_j</math>, bahawa j ganjil adalah nombor Fibonacci ke-(2n).''
:''Jumlah bagi nombor Fibonacci n pertama, <math>F_j</math>, bahawa j genap adalah nombor Fibonacci ke-(2n+1) tolak 1.''

==== Pembuktian ====

Aruhan bagi <math>F_{2n}</math>:
:<math>F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}=F_{2n}</math>
:<math>F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n}+F_{2n+1}</math>
:<math>F_1+F_3+F_5+...+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n+2}</math>
Kes asas ini untuk boleh menjadi <math>F_1=F_2</math>.
<br>
Aruhan bagi <math>F_{2n+1}</math>:
:<math>F_0+F_2+F_4+...+F_{2n-2}+F_{2n}=F_{2n+1}-1</math>
:<math>F_0+F_2+F_4+...+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+1}+F_{2n+2}-1</math>
:<math>F_0+F_2+F_4+...+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+3}-1</math>
Kes asas ini untuk boleh menjadi <math>F_0=F_1-1</math>.

=== Fourth Identity ===

:<math>\sum_{i=0}^n iF_i = nF_{n+2} - F_{n+3} + 2</math>

==== Proof ====

This identity can be established in two stages.
First, we count the number of ways summing 1s and 2s to &minus;1, 0, …, 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''; …; 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
: &nbsp;&nbsp;&nbsp;[''F''(''n'' + 2) &minus; 1] + [''F''(''n'' + 1) &minus; 1] + … + [''F''(2) &minus; 1]
: = [''F''(''n'' + 2) &minus; 1] + [''F''(''n'' + 1) &minus; 1] + … + [''F''(2) &minus; 1] + [''F''(1) &minus; 1] + ''F''(0)
: = ''F''(''n'' + 2) + [''F''(''n'' + 1) + … + ''F''(1) + ''F''(0)] &minus; (''n'' + 2)
: = ''F''(''n'' + 2) + [''F''(''n'' + 3) &minus; 1] &minus; (''n'' + 2)
: = ''F''(''n'' + 2) + ''F''(''n'' + 3) &minus; (''n'' + 3).

On the other hand, we observe from the second identity that there are
* ''F''(0) + ''F''(1) + … + ''F''(''n'' &minus; 1) + ''F''(''n'') ways summing to ''n'' + 1;
* ''F''(0) + ''F''(1) + … + ''F''(''n'' &minus; 1) ways summing to ''n'';
……
* ''F''(0) way summing to &minus;1.
Adding up all ''n'' + 1 sums, we see that there are
* (''n'' + 1) ''F''(0) + ''n'' ''F''(1) + … + ''F''(''n'') ways summing to &minus;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) &minus; (''n'' + 3)

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

=== Fifth Identity ===

:<math>\sum_{i=0}^n {F_i}^2 = F_{n} F_{n+1}</math>

:''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'' ===

:<math>F_{2n} = F_{n+1}^2 - F_{n-1}^2 = F_n(F_{n+1}+F_{n-1}) </math>

<ref name="autogenerated1">[http://mathworld.wolfram.com/FibonacciNumber.html Fibonacci Number - from Wolfram MathWorld<!-- Bot generated title -->]</ref>

=== Another Identity ===
Another identity useful for calculating ''F<sub>n</sub>'' for large values of ''n'' is

:<math>F_{kn+c} = \sum_{i=0}^k {k\choose i} F_{c-i} F_n^i F_{n+1}^{k-i},</math> 
<ref name="autogenerated1" />

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

:<math>F_{2n+k} = F_k F_{n+1}^2 + 2 F_{k-1} F_{n+1} F_n + F_{k-2} F_n^2 </math>

for all integers ''n'' and ''k''. [[Dijkstra]]<ref name="dijkstra78"/> points out that doubling identities of this type can be used to calculate ''F<sub>n</sub>'' 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) <math>{\rm \Theta}(n)\,</math>. Thus the actual performance depends mainly upon efficiency of the implemented [[multiplication algorithm| long multiplication]], and usually is <math>{\rm \Theta}(n \,\log n)</math> or <math>{\rm \Theta}(n ^{\log_2 3})</math>.)

===Other identities===

Other identities include relationships to the [[Lucas number]]s, which have the same recursive properties but start with ''L''<sub>''0''</sub>=2 and ''L''<sub>''1''</sub>=1. These properties include
''F''<sub>''2n''</sub>=''F''<sub>''n''</sub>''L''<sub>''n''</sub>.

There are also scaling identities, which take you from ''F''<sub>n</sub> and ''F''<sub>n+1</sub> to a variety of things of the form ''F''<sub>an+b</sub>; for instance

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

<math>F_{3n+1} = F_{n+1}^3 + 3 F_{n+1}F_n^2 - F_n^3</math>

<math>F_{3n+2} = F_{n+1}^3 + 3 F_{n+1}^2F_n + F_n^3</math>

<math>F_{4n} = 4F_nF_{n+1}(F_{n+1}^2 + 2F_n^2) - 3F_n^2(F_n^2 + 2F_{n+1}^2)</math>

These can be found experimentally using [[lattice reduction]], and are useful in setting up the [[special number field sieve]] to [[Factorization|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 number]]s for details.

==Siri kuasa==
[[Fungsi generasi]] urutan Fibonacci adalah [[siri kuasa]]
:<math>s(x)=\sum_{k=0}^{\infty} F_k x^k.</math>

Siri ini adalah mudah dan jawapan bentuk-tertutup menarik untuk <math>|x| < 1/\varphi</math>
:<math>s(x)=\frac{x}{1-x-x^2}.</math>

Jawapan ini dapat dibukti dengan menggunakan kemunculan semula Fibonacci untuk melebarkan setiap koefisi dalam jumlah infinite mentakrifkan <math>s(x)</math>:
:<math>\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}</math>

Menyelesaikan persamaan <math>s(x)=x+xs(x)+x^2s(x)</math> for <math>s(x)</math> menyebabkan jawapan bentuk tertutup. 

Terutamanya, buku teka-teki matematik menyatakan nilai aneh<math>\frac{s(\frac{1}{10})}{10}=\frac{1}{89}</math>, atau lebih biasanya

:<math>\sum_{n = 1}^{\infty}{\frac {F(n)}{10^{(k + 1)(n + 1)}}} = \frac {1}{10^{2k + 2} - 10^{k + 1} - 1}</math>

untuk semua integer <math>k >= 0</math>.

Secara bicara,
:<math>\sum_{n=0}^\infty\,\frac{F_n}{k^{n}}\,=\,\frac{k}{k^{2}-k-1}.</math>

==Jumlah salingan==

<!--
{{cite book
  | last =Borwein
  | first =Jonathan M.
  | authorlink =Jonathan Borwein
  | coauthors =[[Peter Borwein|Peter B. Borwein]]
  | title =Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity
  | pages =91–101
  | publisher =Wiley
  | year =1998
  | month =July
  | url =http://www.wiley.com/WileyCDA/WileyTitle/productCd-047131515X.html
  | id = ISBN 978-0-471-31515-5 }}
It credits some formulae to {{cite journal | author = Landau, E. | title = Sur la Série des Invers de Nombres de Fibonacci | journal = Bull. Soc. Math. France | volume = 27 | year = 1899 | pages = 298–300}}
-->
Jumlah tidak terhingga ke atas nombor Fibonacci salingan kadang-kadang boleh dinilai dari segi [[fungsi theta]]. Sebagai contoh, kita boleh menulis jumlah setiap nombor Fibonacci salingan indeks ganjil sebagai
:<math>\sum_{k=0}^\infty \frac{1}{F_{2k+1}} = \frac{\sqrt{5}}{4}\vartheta_2^2 \left(0, \frac{3-\sqrt 5}{2}\right) ,</math>

dan jumlah kuasa dua nombor Fibonacci salingan
:<math>\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).</math>

Jika kita menambah 1 untuk setiap nombor Fibonacci dalam jumlah yang pertama, terdapat juga bentuk tertutup
:<math>\sum_{k=0}^\infty \frac{1}{1+F_{2k+1}} = \frac{\sqrt{5}}{2},</math>

dan ada jumlah ''tersarang'' bagi kuasa dua nombor Fibonacci lalu memberi salingan [[nisbah emas]],
:<math>\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2}.</math>

Keputusan seperti ini menjadikannya munasabah bahawa rumus tertutup untuk jumlah mendatar bagi nombor Fibonacci salingan boleh didapati, tetapi tiada yang masih diketahui. Walaupun begitu, [[pemalar Fibonacci salingan]]
:<math>\psi = \sum_{k=1}^{\infty} \frac{1}{F_k} = 3.359885666243 \dots</math> 

telah dibuktikan [[nombor tak nisbah|tidak bernisbah]] oleh [[Richard André-Jeannin]].

==Nombor perdana dan kebolehbahagian==
{{main|Fibonacci perdana}}
'''Fibonacci perdana''' adalah nombor Fibonacci yang [[nombor perdana|perdana ]] {{OEIS|id=A005478}}. The first few are:
: 2, 3, 5, 13, 89, 233, 1597, 28657, 514229, …
Fibonacci perdana dengan beribu-ribu digit telah ditemui, tetapi tidak diketahui sama ada ia terdapat banyak tak terhingga. Semuanya mesti mempunyai indeks utama, kecuali ''F''<sub>4</sub> = 3. Terdapat aturan [[besar sembarangan|yang sewenang-wenangnya panjang]] bagi [[nombor komposit]] dan dengan itu termasuk juga nombor Fibonacci komposit.

Dengan pengecualian 1, 8 dan 144 (''F''<sub>0</sub> = ''F''<sub>1</sub>, ''F''<sub>6</sub> dan ''F''<sub>12</sub>) setiap nombor Fibonacci mempunyai faktor utama yang bukan faktor mana-mana nombor Fibonacci yang lebih kecil ([[teorem Carmichael]]).<ref>Ron Knott, [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibtable.html "The Fibonacci numbers"].</ref>

Tiada nombor Fibonacci lebih besar daripada ''F''<sub>6</sub> = 8 yang lebih besar atau kurang satu daripada nombor perdana.<ref>Ross Honsberger ''Mathematical Gems III'' (AMS Dolciani Mathematcal Expositions No. 9), 1985, ISBN 0-88385-318-3, p. 133.</ref>

Sebarang tiga nombor Fibonacci berturut-turut, yang diambil dua pada satu masa, adalah [[perdana relatif|secara relatifnya perdana]]: itu adalah,
:[[faktor sepunya terbesar|fstb]](''F''<sub>''n''</sub>, ''F''<sub>''n''+1</sub>) = fstb(''F''<sub>''n''</sub>, ''F''<sub>''n''+2</sub>) = 1.
Lebih umum,
:fstb(''F''<sub>''n''</sub>, ''F''<sub>''m''</sub>) = ''F''<sub>fstb(''n'', ''m'').</sub><ref>[[Paulo Ribenboim]], ''My Numbers, My Friends'', Springer-Verlag 2000</ref><ref>Su, Francis E., et al. [http://www.math.hmc.edu/funfacts/ffiles/20004.5.shtml "Fibonacci GCD's, please."], ''Mudd Math Fun Facts''.</ref>

===Pembahagi ganjil===

Jika ''n'' adalah ganjil, semua pembahagi ganjil F<sub>''n''</sub> adalah ≡ 1 (mod 4).<ref>Lemmermeyer, ex. 2.27 p. 73</ref><ref>The website [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibtable.html] has the first 300 Fibonacci numbers factored into primes.</ref> <br>
Ini adalah bersamaan dengan mengatakan bahawa untuk ''n'' ganjil semua faktor perdana ganjil F<sub>''n''</sub> adalah ≡ 1 (mod 4).

<blockquote>Contohnya,
F<sub>1</sub> = 1, F<sub>3</sub> = 2, F<sub>5</sub> = 5, F<sub>7</sub> = 13, F<sub>9</sub> = 34 = 2×17, F<sub>11</sub> = 89, F<sub>13</sub> = 233, F<sub>15</sub> = 610 = 2×5×61
</blockquote>

===Fibonacci dan Legendre===

Terdapat beberapa rumus yang menarik menghubungkan nombor Fibonacci dan [[simbol Legendre]] <math>\;\left(\tfrac{p}{5}\right).</math> 

:<math>
\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.
</math>

Jika ''p'' ialah [[nombor perdana]], maka<ref>[[Paulo Ribenboim]] (1996), ''The New Book of Prime Number Records'', New York: Springer, ISBN 0-387-94457-5, p. 64</ref><ref>Franz Lemmermeyer (2000), ''Reciprocity Laws'', New York: Springer, ISBN 3-540-66957-4, ex 2.25-2.28, pp. 73-74</ref>
<math>
F_{p} \equiv \left(\frac{p}{5}\right) \pmod p \;\;\mbox{ dan }\;\;\;
 
F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p.
</math> 

<blockquote>
Sebagai contoh, 

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

</blockquote>

Juga, jika ''p'' ≠  5 adalah nombor perdana ganjil, maka: <ref>Lemmermeyer, ex. 2.38, pp. 73-74</ref>
:<math>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}
</math>

<blockquote>
Contoh bagi semua kes:

:<math>p=7 \equiv 3 \pmod 4, \;\;(\tfrac{7}{5}) = -1, \frac{5(\frac{7}{5})+3}{2} =-1\mbox{ dan }\frac{5(\frac{7}{5})-3}{2}=-4.</math>  
::<math>F_3=2 \mbox{ dan } F_4=3.</math> 

::<math>5F_3^2=20\equiv -1 \pmod {7}\;\;\mbox{ dan }\;\;5F_4^2=45\equiv -4 \pmod {7}</math>

:<math>p=11 \equiv 3 \pmod 4, \;\;(\tfrac{11}{5}) = +1, \frac{5(\frac{11}{5})+3}{2} =4\mbox{ dan }\frac{5(\frac{11}{5})- 3}{2}=1.</math>  
::<math>F_5=5 \mbox{ dan } F_6=8.</math> 

::<math>5F_5^2=125\equiv 4 \pmod {11} \;\;\mbox{ dan }\;\;5F_6^2=320\equiv 1 \pmod {11}</math>

:<math>p=13 \equiv 1 \pmod 4, \;\;(\tfrac{13}{5}) = -1, \frac{5(\frac{13}{5})-5}{2} =-5\mbox{ dan }\frac{5(\frac{13}{5})+ 5}{2}=0.</math>  
::<math>F_6=8 \mbox{ dan } F_7=13.</math> 

::<math>5F_6^2=320\equiv -5 \pmod {13} \;\;\mbox{ dan }\;\;5F_7^2=845\equiv 0 \pmod {13}</math>

:<math>p=29 \equiv 1 \pmod 4, \;\;(\tfrac{29}{5}) = +1, \frac{5(\frac{29}{5})-5}{2} =0\mbox{ dan }\frac{5(\frac{29}{5})+5}{2}=5.</math>  
::<math>F_{14}=377 \mbox{ dan } F_{15}=610.</math> 

::<math>5F_{14}^2=710645\equiv 0 \pmod {29} \;\;\mbox{ dan }\;\;5F_{15}^2=1860500\equiv 5 \pmod {29}</math>
</blockquote>

===Kebolehbahagian dengan 11===
<math>\sum_{k=n}^{n+9} F_{k} = 11 F_{n+6}</math>

<blockquote>Sebagai contoh, jadi ''n'' = 1:
<br>
F<sub>1</sub>+F<sub>2</sub>+...+F<sub>10</sub> = 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143 = 11×13
<br>
''n'' = 2:
<br>
F<sub>2</sub>+F<sub>3</sub>+...+F<sub>11</sub> = 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 = 231 = 11×21
<br>
''n'' = 3:
<br>
F<sub>3</sub>+F<sub>4</sub>+...+F<sub>12</sub> = 2 +  3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144= 374 = 11×34

</blockquote>

Malah, identitinya adalah benar bagi semua integer ''n'', bukan hanya yang positif:
<blockquote>
''n'' = 0:

<br>
F<sub>0</sub>+F<sub>1</sub>+...+F<sub>9</sub> = 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 88 = 11×8
<br>
''n'' = &minus;1:

<br>
F<sub>&minus;1</sub>+F<sub>0</sub>+...+F<sub>8</sub> = 1 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 = 55 = 11×5
<br>
''n'' = &minus;2:

<br>
F<sub>&minus;2</sub>+F<sub>&minus;1</sub>+F<sub>0</sub>+...+F<sub>7</sub> = &minus;1 + 1 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 = 33 = 11×3

</blockquote>

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

:<math>\displaystyle a_n = F_{2n-1}</math>
:<math>\displaystyle b_n = 2 F_n F_{n-1}</math>
:<math>\displaystyle c_n = {F_n}^2 - {F_{n-1}}^2</math>

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

Any four consecutive Fibonacci numbers ''F''<sub>''n''</sub>, ''F''<sub>''n''+1</sub>, ''F''<sub>''n''+2</sub> and ''F''<sub>''n''+3</sub> can also be used to generate a Pythagorean triple in a different way:
:<math> 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 \,.</math>
Example 1: let the Fibonacci numbers be 1, 2, 3 and 5. Then:
:<math>\displaystyle  a = 1 \times 5 = 5</math>
:<math>\displaystyle  b = 2 \times 2 \times 3 = 12</math>
:<math>\displaystyle  c = 2^2 + 3^2 = 13 \,</math>
:<math>\displaystyle  5^2 + 12^2 = 13^2 \,.</math>
Example 2: let the Fibonacci numbers be 8, 13, 21 and 34. Then:
:<math>\displaystyle  a = 8 \times 34 = 272</math>
:<math>\displaystyle  b = 2 \times 13 \times 21 = 546</math>
:<math>\displaystyle  c = 13^2 + 21^2 = 610 \,</math>
:<math>\displaystyle  272^2 + 546^2 = 610^2 \,.</math>

==Magnitud nombor Fibonacci==
Memandangkan<math>F_n</math> adalah [[berasimptot]] kepada <math>\varphi^n/\sqrt5</math>, bilangan digit dalam asas perwakilan ''b'' <math>F_n\,</math> adalah berasimptot kepada <math>n\,\log_b\varphi</math>.

Dalam asas 10, untuk setiap integer yang lebih besar daripada 1 terdapat 4 atau 5 nombor Fibonacci dengan bilangan digit itu, dalam kebanyakan kes 5.

==Applications==

The Fibonacci numbers are important in the run-time analysis of [[Euclidean algorithm|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 [[Matiyasevich's theorem|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]].<!-- Knuth vol. 2 -->

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.<ref>{{cite journal | author=M. Avriel and D.J. Wilde | title=Optimality of the Symmetric Fibonacci Search Technique | journal=[[Fibonacci Quarterly]] | year=1966 | issue=3 | pages= 265–269}}</ref>

The Fibonacci number series is used for optional [[lossy compression]] in the [[Interchange_File_Format|IFF]] [[8SVX]] audio file format used on [[Amiga]] computers. The number series  [[companding|compands]] the original audio wave similar to logarithmic methods e.g. [[µ-law]].<ref>Amiga ROM Kernel Reference Manual, Addison-Wesley 1991</ref><ref>[http://wiki.multimedia.cx/index.php?title=IFF#Fibonacci_Delta_Compression IFF - MultimediaWiki]</ref>

In [[music]], Fibonacci numbers are sometimes used to determine tunings, and, as in visual art, to determine the length or size of [[content]] or [[form (music)|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 of units|conversion]] factor 1.609344 for [[mile]]s 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 [[Fibonacci coding|number]] [[processor register|register]] in [[golden ratio base]] φ being shifted. To convert from kilometers to miles, shift the register down the Fibonacci sequence instead.<ref>[http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrep.html#kilos An Application of the Fibonacci Number Representation]</ref><ref>[http://people.bath.ac.uk/pst20/fibonacci.html#Sequence A Practical Use of the Sequence]</ref><ref>[http://eom.springer.de/Z/z120020.htm Zeckendorf representation]</ref>

==Fibonacci numbers in nature==
[[Image:Helianthus whorl.jpg|thumb|[[Sunflower]] head displaying florets in spirals of 34 and 55 around the outside]]
Fibonacci sequences appear in biological settings,<ref>{{cite journal | author=S. Douady and Y. Couder | title=Phyllotaxis as a Dynamical Self Organizing Process | journal=Journal of Theoretical Biology | year=1996 | issue=178 | pages= 255&ndash;274 | url=http://www.math.ntnu.no/~jarlet/Douady96.pdf  | doi = 10.1006/jtbi.1996.0026 | volume=178|format=PDF}}</ref> in two consecutive Fibonacci numbers, such as branching in trees, arrangement of [[leaves]] on a stem, the fruitlets of a [[pineapple]],<ref>{{cite book|first=Judy|last=Jones|coauthors=William Wilson|title=An Incomplete Education|publisher=Ballantine Books|year=2006|id=ISBN 978-0-7394-7582-9|pages=544|chapter=Science}}</ref> the flowering of [[artichoke]], an uncurling fern and the arrangement of  a [[pine cone]].<ref>{{cite journal | author=A. Brousseau | title=Fibonacci Statistics in Conifers | journal=[[Fibonacci Quarterly]] | year=1969 | issue=7 | pages= 525–532}}</ref> In addition, numerous poorly substantiated claims of Fibonacci numbers or [[golden section]]s in nature are found in popular sources, e.g. relating to the breeding of rabbits, the spirals of shells, and the curve of waves{{Fact|date=February 2007}}.  The Fibonacci numbers are also found in the family tree of honeybees. <ref>[http://www.cs4fn.org/maths/bee-davinci.php Computer Science for Fun - cs4fn: Marks for the da Vinci Code: B<!-- Bot generated title -->]</ref>

[[Przemyslaw Prusinkiewicz]] advanced the idea that real instances can be in part understood as the expression of certain algebraic constraints on [[free group]]s, specifically as certain [[L-system|Lindenmayer grammar]]s.<ref>{{cite book|first=Przemyslaw|last=Prusinkiewicz|coauthors=James Hanan|title=Lindenmayer Systems, Fractals, and Plants (Lecture Notes in Biomathematics)|publisher=[[Springer Science+Business Media|Springer-Verlag]]|year=1989|id=ISBN 0-387-97092-4}}</ref>

A model for the pattern of [[floret]]s in the head of a [[sunflower]] was proposed by H. Vogel in 1979.<ref>
{{Citation
  | last =Vogel
  | first =H
  | title =A better way to construct the sunflower head
  | journal =Mathematical Biosciences
  | issue =44
  | pages =179–189
  | year =1979
  | doi =10.1016/0025-5564(79)90080-4
  | volume =44
}}</ref>
This has the form
:<math>\theta = \frac{2\pi}{\phi^2} n</math>, <math>r = c \sqrt{n}</math>
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.<ref>{{cite book
  | last =Prusinkiewicz
  | first =Przemyslaw
  | authorlink =Przemyslaw Prusinkiewicz
  | coauthors =[[Aristid Lindenmayer|Lindenmayer, Aristid]]
  | title =[[The Algorithmic Beauty of Plants]]
  | publisher =Springer-Verlag
  | year= 1990
  | location =
  | pages =101-107
  | url =http://algorithmicbotany.org/papers/#webdocs
  | doi =
  | id = ISBN 978-0387972978 }}</ref>
* In 1991, Jean-Claude Perez proposed a connection between [[DNA]] [[base sequence]]s within [[gene sequence]]s and Fibonacci Numbers <ref>J.C. Perez (1991), [http://golden-ratio-in-dna.blogspot.com/2008/01/1991-first-publication-related-to.html "Chaos DNA and Neuro-computers: A Golden Link"], in ''Speculations in Science and Technology'' vol. 14 no. 4, {{ISSN|0155-7785}}</ref>.

== Budaya popular ==
{{main|Nombor Fibonacci dalam budaya popular}}
<!--NOTE: YOUR FAVOURITE FIBONACCI REFERENCE SHOULD ONLY BE IN MAIN ARTICLE (Fibonacci numbers in popular culture) AND MAY ALREADY BE THERE!-->

==Generalizations==
{{main|Generalizations of Fibonacci numbers}}
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]]. <ref>{{MathWorld|title=Fibonacci Number|urlname=FibonacciNumber|author=Pravin Chandra and [[Eric W. Weisstein]]}}</ref>
* Starting with other integers. [[Lucas number]]s have ''L''<sub>1</sub> = 1, ''L''<sub>2</sub> = 3, and ''L<sub>n</sub>'' = ''L''<sub>''n''−1</sub> + ''L''<sub>''n''−2</sub>. [[Primefree sequence]]s use the Fibonacci recursion with other starting points in order to generate sequences in which all numbers are [[composite number|composite]].
* Letting a number be a linear function (other than the sum) of the 2 preceding numbers. The [[Pell number]]s have ''P<sub>n</sub>'' = 2''P''<sub>''n'' – 1</sub> + ''P''<sub>''n'' – 2</sub>.
* Not adding the immediately preceding numbers. The [[Padovan sequence]] and [[Perrin number]]s 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 sequence|periodic]] with period at most <math>n^2</math>.  The lengths of the periods for various ''n'' form the so-called [[Pisano period]]s {{OEIS|id=A001175}}.  Determining the Pisano periods in general is an open problem,{{Fact|date=March 2008}} 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.<ref>[http://american-university.com/cas/mathstat/newstudents/shared/puzzles/fibbee.html The Fibonacci Numbers and the Ancestry of Bees]</ref>

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.<ref>{{cite article | title=Square Fibonacci Numbers Etc |author= J H E Cohn |journal= Fibonacci Quarterly | volume= 2 | year= 1964 | pages=109-113 | url= http://math.la.asu.edu/~checkman/SquareFibonacci.html}}</ref>

In 1990, Jean-claude Perez published strong links between [[fractals]] world and Fibonacci numbers sensitivity <ref>[[IEEE]] [http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel2/148/3745/00137678.pdf?arnumber=137678 Integers neural network systems (INNS) using resonance propertiesof a Fibonacciapos;s chaotic `golden neuronapos] 1990</ref><ref>[http://golden-ratio-in-dna.blogspot.com/2008/01/1992-order-and-chaos-in-dnathe-denis.html Golden ratio and numbers in DNA] 2008</ref>

==Lihat pula==
*[[Logarithmic spiral]]
*[[b:Atur cara bilangan Fibonacci|Atur cara bilangan Fibonacci]] di [[Bukuwiki]]
*[[Pertubuhan Fibonacci]]
*[[Fibonacci Quarterly]] &mdash; sebuah jurnal akademik devoted pada kajian nombor Fibonacci
*Bilangan [[Negafibonacci]]
*[[Bilangan Lucas]]

==References==
{{reflist|2}}

==External links==
{{external links}}
* Peter Marcer, ''[http://golden-ratio-in-dna.blogspot.com/2008/01/1992-order-and-chaos-in-dnathe-denis.html describing the discovery by jean-claude Perez of Fibonacci numbers structuring proportions of TCAG nucleotides within DNA]'', (1992).
* Ron Knott, ''[http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi.html The Golden Section: Phi]'', (2005).
* Ron Knott, ''[http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrep.html Representations of Integers using Fibonacci numbers]'', (2004).
* wallstreetcosmos.com, ''[http://www.wallstreetcosmos.com/elliot.html Fibonacci numbers and stock market analysis]'', (2008).
* Juanita Lofthouse ''[http://arxiv.org/abs/physics/0411169 Fibonacci numbers and Red Blood Cell Dynamics]'', .
* Bob Johnson, ''[http://www.dur.ac.uk/bob.johnson/fibonacci/ Fibonacci resources]'', (2004)
* Donald E. Simanek, ''[http://www.lhup.edu/~dsimanek/pseudo/fibonacc.htm Fibonacci Flim-Flam]'', (undated, 2005 or earlier).
* Rachel Hall, ''[http://www.sju.edu/~rhall/Multi/rhythm2.pdf Hemachandra's application to Sanskrit poetry]'', (undated; 2005 or earlier).
* Alex Vinokur, ''[http://semillon.wpi.edu/~aofa/AofA/msg00012.html Computing Fibonacci numbers on a Turing Machine]'', (2003).
* (no author given), ''[http://www.goldenmeangauge.co.uk/fibonacci.htm Fibonacci Numbers Information]'', (undated, 2005 or earlier).
* [http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html Fibonacci Numbers and the Golden Section] – Ron Knott's Surrey University multimedia web site on the Fibonacci numbers, the Golden section and the Golden string. 
* The [http://www.mscs.dal.ca/Fibonacci/ Fibonacci Association] incorporated in [[1963]], focuses on Fibonacci numbers and related mathematics, emphasizing new results, research proposals, challenging problems, and new proofs of old ideas.
* Dawson Merrill's [http://www.goldenratio.org/info/ Fib-Phi] link page.
* [http://primes.utm.edu/glossary/page.php?sort=FibonacciPrime Fibonacci primes]
* [http://www.mathpages.com/home/kmath078.htm Periods of Fibonacci Sequences Mod m] at MathPages
* [http://www.upl.cs.wisc.edu/~bethenco/fibo/ The One Millionth Fibonacci Number]
* [http://www.bigzaphod.org/fibonacci/ The Ten Millionth Fibonacci Number]
* An [http://www.calcresult.com/maths/Sequences/expanded_fibonacci.html Expanded Fibonacci Series Generator]
* Manolis Lourakis, [http://www.ics.forth.gr/~lourakis/fibsrch/ Fibonaccian search in C]
* [http://www.physorg.com/news97227410.html Scientists find clues to the formation of Fibonacci spirals in nature]
*[http://web.archive.org/web/20070715032716/http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=630&bodyId=1002 Fibonacci Numbers] at [http://web.archive.org/web/20060212072618/http://mathdl.maa.org/convergence/1/ Convergence]
* [http://www.tools4noobs.com/online_tools/fibonacci/ Online Fibonacci calculator]

[[Kategori:Fibonacci numbers|*]]
[[Kategori:Articles containing proofs]]

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