Nombor Fibonacci

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(Dilencongkan dari Bilangan Fibonacci)
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Suatu ubinan dengan segi empat yang tepinya adalah nombor Fibonaci berturut-turut pada panjangnya
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.[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, 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.[2][3]

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:


  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 nombor adalah jumlah dua nombor yang terdahulu. Nombor Fibonacci pertama (jujukan A000045 dalam 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.

Asal Usul[sunting | sunting sumber]

Nombor Fibonacci pertama kali muncul, di bawah nama mātrāmeru (gunung irama), dalam karya ahli tatabahasa Pingala (Chandah-shāstra, Seni Prosodi, 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.

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: 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 − 1, atau L kepada corak panjang n − 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 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)[6]. Dia menganggap pertumbuhan unggul populasi arnab (secara biologinya tidak realistik), dengan anggapan bahawa:

  • Dalam bulan "sifar", ada sepasang arnab (pasangan tambahan arnab = 0)
  • Dalam bulan pertama, pasangan pertama beranak sepasang lagi (pasangan tambahan arnab = 1)
  • Dalam bulan kedua, kedua-dua pasang arnab mempunyai sepasang lagi, dan pasangan yang pertama mati (pasangan tambahan arnab = 1)
  • Dalam bulan ketiga, pasangan kedua dan kedua-dua pasangan baru mempunyai sejumlah tiga pasangan baru, dan pasangan kedua tua mati. (pasangan tambahan arnab = 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 − 2 adalah subur dan melahirkan anak, jadi F(n − 2) pasangan ditambah kepada populasi semasa F(n − 1). Oleh itu, jumlah F(n) = F(n − 1) + F(n − 2).[7]

Kaitannya dengan Nisbah Emas[sunting | sunting sumber]

Closed form expression[sunting | sunting sumber]

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\, (jujukan A001622 dalam 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[sunting | sunting sumber]

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.

Pengiraan melalui pembundaran[sunting | sunting sumber]

Memandangkan \begin{matrix}|1-\varphi|^n/\sqrt 5 < 1/2\end{matrix} bagi semua n\geq 0, nombor F(n) adalah integer yang paling hampir dengan \varphi^n/\sqrt 5\, . Oleh itu, ia boleh didapati dengan pembundaran, atau dari segi fungsi lantai:

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

Limit of consecutive quotients[sunting | sunting sumber]

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[sunting | sunting sumber]

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[sunting | sunting sumber]

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[sunting | sunting sumber]

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]

Pengenalan[sunting | sunting sumber]

Kebanyakan pengenalan melibatkan nombor Fibonacci menarik dari hujah kombinatorik. F(n) boleh ditafsirkan sebagai bilangan cara menjumlahkan 1 dan 2 kepada n − 1, dengan kelaziman yang F(0) = 0, bermakna tiada jumlah akan menambah sehingga −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[sunting | sunting sumber]

F_{n+1} = F_{n} + F_{n-1}
Nombor Fibonacci ke-n adalah jumlah dua nombor Fibonacci sebelumnya.

Pembuktian[sunting | sunting sumber]

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−1, maka F(n) menjumlah pada set pertama. Set kedua yang mengandungi jumlah penghasil tambah pertamanya 2; jumlah baki kepada n−2, maka F(n−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−1).

Pengenalan Kedua[sunting | sunting sumber]

\sum_{i=0}^n F_i = F_{n+2} - 1
Jumlah bagi n pertama nombor Fibonacci adalah nombor Fibonacci ke-(n+2) tolak 1.

Pembuktian[sunting | sunting sumber]

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 − 1. Jika penghasil tambah yang kedua ialah 2 tetapi yang pertama adalah 1, maka terdapat F(n − 1) cara untuk melengkapkan pengiraan bagi n − 2. Teruskan dengan cara ini. Kemudiannya kita mempertimbangkan penghasil tambah ke-(n + 1). Jika ia adalah 2 tetapi semua penghasil tambahn 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 − 1) + … + F(0) memberikan pengiraan yang dikehendaki.

Pengenalan Ketiga[sunting | sunting sumber]

Identiti ini mempunyai bentuk yang sedikit berbeza untuk F_k, bergantung kepada sama ada k adalah ganjil atau genap.

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

[12]

Jumlah bagi nombor Fibonacci n-1 pertama, F_j, supaya j ganjil adalah nombor Fibonacci ke-(2n).
Jumlah bagi nombor Fibonacci n pertama, F_j, supaya j genap adalah nombor Fibonacci ke-(2n+1) tolak 1.

Pembuktian[sunting | sunting sumber]

Aruhan bagi 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}

Kes asas ini untuk boleh menjadi F_1=F_2.
Aruhan bagi 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

Kes asas ini untuk boleh menjadi F_0=F_1-1.

Pengenalan Keempat[sunting | sunting sumber]

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

Pembuktian[sunting | sunting sumber]

Pengenalan ini boleh ditentukan dalam dua peringkat. Pertama, kita mengira bilangan cara menjumlahkan 1 dan 2 menjadi −1, 0, …, atau n + 1 supaya sekurang-kurangnya salah satu daripada penghasil tambah adalah 2.

Dengan pengenalan kedua kita, terdapat F(n + 2) − 1 cara menjumlahkan kepada n + 1; F(n + 1) − 1 cara menjumlahkan kepada n; …; dan, akhirnya, F(2) − 1 cara menjumlahkan kepada 1. Apabila F(1) − 1 = F(0) = 0, kita boleh menambah semua jumlah n + 1 dan menggunakan pengenalan kedua lagi untuk mendapatkan:    [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).

Sebaliknya, kita dapati daripada pengenalan kedua bahawa terdapat

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

……

  • F(0) cara menjumlahkan kepada −1.

Menambahkan semua jumlah n + 1, kita dapat melihat bahawa terdapat

  • (n + 1) F(0) + n F(1) + … + F(n) cara menjumlahkan kepada −1, 0, …, or n + 1.

Memandangkan kedua-dua kaedah pengiraan merujuk kepada nombor yang sama, kita dapat

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

Akhir sekali, kita melengkapkan bukti dengan membuat tolakan pengenalan di atas daripada n + 1 kali pengenalan kedua.

Pengenalan Kelima[sunting | sunting sumber]

\sum_{i=0}^n {F_i}^2 = F_{n} F_{n+1}
Jumlah bagi nombor Fibonacci n yang pertama dikuasa dua adalah hasil darab nombor Fibonacci ke-n dan ke-(n+1).

Pengenalan bagi n berganda[sunting | sunting sumber]

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

[13]

Another Identity[sunting | sunting sumber]

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[sunting | sunting sumber]

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.

Siri kuasa[sunting | sunting sumber]

Fungsi generasi urutan Fibonacci adalah siri kuasa

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 teka-teki matematik menyatakan nilai aneh\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}.

Jumlah salingan[sunting | sunting sumber]

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

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

dan jumlah kuasa dua nombor Fibonacci salingan

\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).

Jika kita menambah 1 untuk setiap nombor Fibonacci dalam jumlah yang pertama, terdapat juga bentuk tertutup

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

dan ada jumlah tersarang bagi kuasa dua nombor Fibonacci lalu memberi salingan nisbah emas,

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

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

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

telah dibuktikan tidak bernisbah oleh Richard André-Jeannin.

Nombor perdana dan kebolehbahagian[sunting | sunting sumber]

Rencana utama: Fibonacci perdana

Fibonacci perdana adalah nombor Fibonacci yang perdana (jujukan A005478 dalam OEIS). 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 F4 = 3. Terdapat aturan yang sewenang-wenangnya panjang bagi nombor komposit dan dengan itu termasuk juga nombor Fibonacci komposit.

Dengan pengecualian 1, 8 dan 144 (F0 = F1, F6 dan F12) setiap nombor Fibonacci mempunyai faktor utama yang bukan faktor mana-mana nombor Fibonacci yang lebih kecil (teorem Carmichael).[14]

Tiada nombor Fibonacci lebih besar daripada F6 = 8 yang lebih besar atau kurang satu daripada nombor perdana.[15]

Sebarang tiga nombor Fibonacci berturut-turut, yang diambil dua pada satu masa, adalah secara relatifnya perdana: itu adalah,

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

Lebih umum,

fstb(Fn, Fm) = Ffstb(n, m).[16][17]

Pembahagi ganjil[sunting | sunting sumber]

Jika n adalah ganjil, semua pembahagi ganjil Fn adalah ≡ 1 (mod 4).[18][19]
Ini adalah bersamaan dengan mengatakan bahawa untuk n ganjil semua faktor perdana ganjil Fn adalah ≡ 1 (mod 4).

Contohnya,

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

Fibonacci dan Legendre[sunting | sunting sumber]

Terdapat beberapa rumus yang menarik menghubungkan nombor Fibonacci dan simbol Legendre \;\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.

Jika p ialah nombor perdana, maka[20][21] 
F_{p} \equiv \left(\frac{p}{5}\right) \pmod p \;\;\mbox{ dan }\;\;\;
 
F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p.

Sebagai contoh,

(\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.

Juga, jika p ≠ 5 adalah nombor perdana ganjil, maka: [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}

Contoh bagi semua kes:

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

Kebolehbahagian dengan 11[sunting | sunting sumber]

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

Sebagai contoh, jadi 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

Malah, identitinya adalah benar bagi semua integer n, bukan hanya yang positif:

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[sunting | sunting sumber]

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 \,.

Magnitud nombor Fibonacci[sunting | sunting sumber]

MemandangkanF_n adalah berasimptot kepada \varphi^n/\sqrt5, bilangan digit dalam asas perwakilan b F_n\, adalah berasimptot kepada n\,\log_b\varphi.

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[sunting | sunting sumber]

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[sunting | sunting sumber]

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]

Budaya popular[sunting | sunting sumber]

Generalizations[sunting | sunting sumber]

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[sunting | sunting sumber]

Periodicity mod n: Pisano periods[sunting | sunting sumber]

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 (jujukan A001175 dalam 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[sunting | sunting sumber]

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[sunting | sunting sumber]

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]

Lihat pula[sunting | sunting sumber]

References[sunting | sunting sumber]

  1. http://www.quipus.it/english/Andean%20Calculators.pdf
  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.
  5. The website [1] has the first 300 Fn factored into primes and links to more extensive tables.
  6. Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. ISBN 0-387-95419-8.  Chapter II.12, pp. 404–405.
  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. p. 92. ISBN 0198581203.  Strena seu de Nive Sexangula (1611)
  9. 9.0 9.1 E. W. Dijkstra (1978). In honour of Fibonacci. Report EWD654
  10. Posamentier, Alfred; Lehmann, Ingmar (2007). The (Fabulous) FIBONACCI Numbers. Prometheus Books. p. 305. ISBN 978-1-59102-475-0. 
  11. M. Möbius, Wie erkennt man eine Fibonacci Zahl?, Math. Semesterber. (1998) 45; 243–246
  12. Vorobiev, Nikolaĭ Nikolaevich; Mircea Martin (2002). "Chapter 1". Fibonacci Numbers. Birkhäuser. pp. pp. 5–6. ISBN 3-7643-6135-2. 
  13. 13.0 13.1 Fibonacci Number - from Wolfram MathWorld
  14. Ron Knott, "The Fibonacci numbers".
  15. Ross Honsberger Mathematical Gems III (AMS Dolciani Mathematcal Expositions No. 9), 1985, ISBN 0-88385-318-3, p. 133.
  16. Paulo Ribenboim, My Numbers, My Friends, Springer-Verlag 2000
  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. p. 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. Prusinkiewicz, Przemyslaw; Lindenmayer, Aristid (1990). [[The Algorithmic Beauty of Plants]]. Springer-Verlag. pp. 101–107. ISBN 978-0387972978.  Wikilink embedded in URL title (bantuan)
  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 (1964). "Square Fibonacci Numbers Etc". Fibonacci Quarterly 2. pp. 109–113. 
  40. IEEE Integers neural network systems (INNS) using resonance propertiesof a Fibonacciapos;s chaotic `golden neuronapos 1990
  41. Golden ratio and numbers in DNA 2008

External links[sunting | sunting sumber]