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Fibonacci sequence

From Wikipedia, the free encyclopedia
Numbers obtained by adding the two previous ones
For the chamber ensemble, seeFibonacci Sequence (ensemble).

In mathematics, theFibonacci sequence is asequence in which each element is the sum of the two elements that precede it. Numbers that are part of the Fibonacci sequence are known asFibonacci numbers, commonly denotedFn . Many writers begin the sequence with 0 and 1, although some authors start it from 1 and 1[1][2] and some (as did Fibonacci) from 1 and 2. Starting from 0 and 1, the sequence begins

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... (sequenceA000045 in theOEIS)
A tiling withsquares whose side lengths are successive Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13 and 21

The Fibonacci numbers were first described inIndian mathematics as early as 200 BC in work byPingala on enumerating possible patterns ofSanskrit poetry formed from syllables of two lengths.[3][4][5] They are named after the Italian mathematician Leonardo of Pisa, also known asFibonacci, who introduced the sequence to Western European mathematics in his 1202 bookLiber Abaci.[6]

Fibonacci numbers appear unexpectedly often in mathematics, so much so that there is an entire journal dedicated to their study, theFibonacci Quarterly. Applications of Fibonacci numbers include computer algorithms such as theFibonacci search technique and theFibonacci heapdata structure, andgraphs calledFibonacci cubes used for interconnecting parallel and distributed systems. They also appearin biological settings, such as branching in trees,the arrangement of leaves on a stem, the fruit sprouts of apineapple, the flowering of anartichoke, and the arrangement of apine cone's bracts, though they do not occur in all species.

Fibonacci numbers are also strongly related to thegolden ratio:Binet's formula expresses then-th Fibonacci number in terms ofn and the golden ratio, and implies that the ratio of two consecutive Fibonacci numbers tends to the golden ratio asn increases. Fibonacci numbers are also closely related toLucas numbers, which obey the samerecurrence relation and with the Fibonacci numbers form a complementary pair ofLucas sequences.

Definition

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The Fibonacci spiral: an approximation of thegolden spiral created by drawingcircular arcs connecting the opposite corners of squares in the Fibonacci tiling (see preceding image)

The Fibonacci numbers may be defined by therecurrence relation[7]F0=0,F1=1,{\displaystyle F_{0}=0,\quad F_{1}=1,}andFn=Fn−1+Fn−2{\displaystyle F_{n}=F_{n-1}+F_{n-2}}forn > 1.

Under some older definitions, the valueF0=0{\displaystyle F_{0}=0} is omitted, so that the sequence starts withF1=F2=1,{\displaystyle F_{1}=F_{2}=1,} and the recurrenceFn=Fn−1+Fn−2{\displaystyle F_{n}=F_{n-1}+F_{n-2}} is valid forn > 2.[8][9]

The first 20 Fibonacci numbersFn are:

F0F1F2F3F4F5F6F7F8F9F10F11F12F13F14F15F16F17F18F19
01123581321345589144233377610987159725844181

History

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India

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See also:Golden ratio § History
Thirteen (F7) ways of arranging long and short syllables in a cadence of length six. Eight (F6) end with a short syllable and five (F5) end with a long syllable.

The Fibonacci sequence appears inIndian mathematics, in connection withSanskrit prosody.[4][10][11] In the Sanskrit poetic tradition, there was interest in enumerating all patterns of long (L) syllables of 2 units duration, juxtaposed with short (S) syllables of 1 unit duration. Counting the different patterns of successive L and S with a given total duration results in the Fibonacci numbers: the number of patterns of durationm units isFm+1.[5]

Knowledge of the Fibonacci sequence was expressed as early asPingala (c. 450 BC–200 BC). Singh cites Pingala's cryptic formulamisrau cha ("the two are mixed") and scholars who interpret it in context as saying that the number of patterns form beats (Fm+1) is obtained by adding one [S] to theFm cases and one [L] to theFm−1 cases.[12]Bharata Muni also expresses knowledge of the sequence in theNatya Shastra (c. 100 BC–c. 350 AD).[4][3]However, the clearest exposition of the sequence arises in the work ofVirahanka (c. 700 AD), whose own work is lost, but is available in a quotation by Gopala (c. 1135):[11]

Variations of two earlier meters [is the variation] ... For example, for [a meter of length] four, variations of meters of two [and] three being mixed, five happens. [works out examples 8, 13, 21] ... In this way, the process should be followed in allmātrā-vṛttas [prosodic combinations].[a]

Hemachandra (c. 1150) is credited with knowledge of the sequence as well,[3] writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."[14][15]

Europe

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A page ofFibonacci'sLiber Abaci from theBiblioteca Nazionale di Firenze showing (in box on right) 13 entries of the Fibonacci sequence:
the indices from present to XII (months) as Latin ordinals and Roman numerals and the numbers (of rabbit pairs) as Hindu-Arabic numerals starting with 1, 2, 3, 5 and ending with 377.

The Fibonacci sequence first appears in the bookLiber Abaci (The Book of Calculation, 1202) byFibonacci,[16][17] where it is used to calculate the growth of rabbit populations.[18][19] Fibonacci considers the growth of an idealized (biologically unrealistic)rabbit population, assuming that: a newly born breeding pair of rabbits are put in a field; each breeding pair mates at the age of one month, and at the end of their second month they always produce another pair of rabbits; and rabbits never die, but continue breeding forever. Fibonacci posed the rabbitmath problem: how many pairs will there be in one year?

  • At the end of the first month, they mate, but there is still only 1 pair.
  • At the end of the second month they produce a new pair, so there are 2 pairs in the field.
  • At the end of the third month, the original pair produce a second pair, but the second pair only mate to gestate for a month, so there are 3 pairs in all.
  • At the end of the fourth month, the original pair has produced yet another new pair, and the pair born two months ago also produces their first pair, making 5 pairs.

At the end of then-th month, the number of pairs of rabbits is equal to the number of mature pairs (that is, the number of pairs in monthn – 2) plus the number of pairs alive last month (monthn – 1). The number in then-th month is then-th Fibonacci number.[20]

The name "Fibonacci sequence" was first used by the 19th-century number theoristÉdouard Lucas.[21]

Solution to Fibonacci rabbitproblem: In a growing idealized population, the number of rabbit pairs form the Fibonacci sequence. Atthe end of the nth month, the number of pairs is equal toFn.

Relation to the golden ratio

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Main article:Golden ratio

Closed-form expression

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Like everysequence defined by a homogeneouslinear recurrence with constant coefficients, the Fibonacci numbers have aclosed-form expression.[22] It has become known asBinet's formula, named after French mathematicianJacques Philippe Marie Binet, though it was already known byAbraham de Moivre andDaniel Bernoulli:[23]

Fn=φn−ψnφ−ψ=φn−ψn5,{\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}={\frac {\varphi ^{n}-\psi ^{n}}{\sqrt {5}}},}

where

φ=1+52≈1.6180339887…{\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803\,39887\ldots }

is thegolden ratio, andψ{\displaystyle \psi } is itsconjugate:[24]

ψ=1−52=1−φ=−1φ≈−0.6180339887….{\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}=1-\varphi =-{1 \over \varphi }\approx -0.61803\,39887\ldots .}

Sinceψ=−φ−1{\displaystyle \psi =-\varphi ^{-1}}, this formula can also be written as

Fn=φn−(−φ)−n5=φn−(−φ)−n2φ−1.{\displaystyle F_{n}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{2\varphi -1}}.}

To see the relation between the sequence and these constants,[25] note thatφ{\displaystyle \varphi } andψ{\displaystyle \psi } are both solutions of the equationx2=x+1{\textstyle x^{2}=x+1} and thusxn=xn−1+xn−2,{\displaystyle x^{n}=x^{n-1}+x^{n-2},} so the powers ofφ{\displaystyle \varphi } andψ{\displaystyle \psi } satisfy the Fibonacci recursion. In other words,

φn=φn−1+φn−2,ψn=ψn−1+ψn−2.{\displaystyle {\begin{aligned}\varphi ^{n}&=\varphi ^{n-1}+\varphi ^{n-2},\\[3mu]\psi ^{n}&=\psi ^{n-1}+\psi ^{n-2}.\end{aligned}}}

It follows that for any valuesa andb, the sequence defined by

Un=aφn+bψn{\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}

satisfies the same recurrence,

Un=aφn+bψn=a(φn−1+φn−2)+b(ψn−1+ψn−2)=aφn−1+bψn−1+aφn−2+bψn−2=Un−1+Un−2.{\displaystyle {\begin{aligned}U_{n}&=a\varphi ^{n}+b\psi ^{n}\\[3mu]&=a(\varphi ^{n-1}+\varphi ^{n-2})+b(\psi ^{n-1}+\psi ^{n-2})\\[3mu]&=a\varphi ^{n-1}+b\psi ^{n-1}+a\varphi ^{n-2}+b\psi ^{n-2}\\[3mu]&=U_{n-1}+U_{n-2}.\end{aligned}}}

Ifa andb are chosen so thatU0 = 0 andU1 = 1 then the resulting sequenceUn must be the Fibonacci sequence. This is the same as requiringa andb satisfy the system of equations:

{a+b=0φa+ψb=1{\displaystyle \left\{{\begin{aligned}a+b&=0\\\varphi a+\psi b&=1\end{aligned}}\right.}

which has solution

a=1φ−ψ=15,b=−a,{\displaystyle a={\frac {1}{\varphi -\psi }}={\frac {1}{\sqrt {5}}},\quad b=-a,}

producing the required formula.

Taking the starting valuesU0 andU1 to be arbitrary constants, a more general solution is:

Un=aφn+bψn{\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}

where

a=U1−U0ψ5,b=U0φ−U15.{\displaystyle {\begin{aligned}a&={\frac {U_{1}-U_{0}\psi }{\sqrt {5}}},\\[3mu]b&={\frac {U_{0}\varphi -U_{1}}{\sqrt {5}}}.\end{aligned}}}

Computation by rounding

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Since|ψn5|<12{\textstyle \left|{\frac {\psi ^{n}}{\sqrt {5}}}\right|<{\frac {1}{2}}} for alln ≥ 0, the numberFn is the closestinteger toφn5{\displaystyle {\frac {\varphi ^{n}}{\sqrt {5}}}}. Therefore, it can be found byrounding, using the nearest integer function:Fn=⌊φn5⌉, n≥0.{\displaystyle F_{n}=\left\lfloor {\frac {\varphi ^{n}}{\sqrt {5}}}\right\rceil ,\ n\geq 0.}

In fact, the rounding error quickly becomes very small asn grows, being less than 0.1 forn ≥ 4, and less than 0.01 forn ≥ 8. This formula is easily inverted to find an index of a Fibonacci numberF:n(F)=⌊logφ⁡5F⌉, F≥1.{\displaystyle n(F)=\left\lfloor \log _{\varphi }{\sqrt {5}}F\right\rceil ,\ F\geq 1.}

Instead using thefloor function gives the largest index of a Fibonacci number that is not greater thanF:nlargest(F)=⌊logφ⁡5(F+1/2)⌋, F≥0,{\displaystyle n_{\mathrm {largest} }(F)=\left\lfloor \log _{\varphi }{\sqrt {5}}(F+1/2)\right\rfloor ,\ F\geq 0,}wherelogφ⁡(x)=ln⁡(x)/ln⁡(φ)=log10⁡(x)/log10⁡(φ){\displaystyle \log _{\varphi }(x)=\ln(x)/\ln(\varphi )=\log _{10}(x)/\log _{10}(\varphi )},ln⁡(φ)=0.481211…{\displaystyle \ln(\varphi )=0.481211\ldots },[26] andlog10⁡(φ)=0.208987…{\displaystyle \log _{10}(\varphi )=0.208987\ldots }.[27]

Magnitude

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SinceFn isasymptotic toφn/5{\displaystyle \varphi ^{n}/{\sqrt {5}}}, the number of digits inFn is asymptotic tonlog10⁡φ≈0.2090n{\displaystyle n\log _{10}\varphi \approx 0.2090\,n}. As a consequence, for every integerd > 1 there are either 4 or 5 Fibonacci numbers withd decimal digits.

More generally, in thebaseb representation, the number of digits inFn is asymptotic tonlogb⁡φ=nlog⁡φlog⁡b.{\displaystyle n\log _{b}\varphi ={\frac {n\log \varphi }{\log b}}.}

Limit of consecutive quotients

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Johannes Kepler observed that the ratio of consecutive Fibonacci numbersconverges. 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 these ratios approach the golden ratioφ:{\displaystyle \varphi \colon }[28][29]limn→∞Fn+1Fn=φ.{\displaystyle \lim _{n\to \infty }{\frac {F_{n+1}}{F_{n}}}=\varphi .}

This convergence holds regardless of the starting valuesU0{\displaystyle U_{0}} andU1{\displaystyle U_{1}}, unlessU1=−U0/φ{\displaystyle U_{1}=-U_{0}/\varphi }. This can be verified usingBinet's formula. For example, the initial values 3 and 2 generate the sequence 3, 2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, ... . The ratio of consecutive elements in this sequence shows the same convergence towards the golden ratio.

In general,limn→∞Fn+mFn=φm{\displaystyle \lim _{n\to \infty }{\frac {F_{n+m}}{F_{n}}}=\varphi ^{m}}, because the ratios between consecutive Fibonacci numbers approachesφ{\displaystyle \varphi }.

Successive tilings of the plane and a graph of approximations to the golden ratio calculated by dividing each Fibonacci number by the previous

Decomposition of powers

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Since the golden ratio satisfies the equationφ2=φ+1,{\displaystyle \varphi ^{2}=\varphi +1,}

this expression can be used to decompose higher powersφn{\displaystyle \varphi ^{n}} as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination ofφ{\displaystyle \varphi } and 1. The resultingrecurrence relationships yield Fibonacci numbers as the linearcoefficients:φn=Fnφ+Fn−1.{\displaystyle \varphi ^{n}=F_{n}\varphi +F_{n-1}.}This equation can beproved byinduction onn ≥ 1:φn+1=(Fnφ+Fn−1)φ=Fnφ2+Fn−1φ=Fn(φ+1)+Fn−1φ=(Fn+Fn−1)φ+Fn=Fn+1φ+Fn.{\displaystyle \varphi ^{n+1}=(F_{n}\varphi +F_{n-1})\varphi =F_{n}\varphi ^{2}+F_{n-1}\varphi =F_{n}(\varphi +1)+F_{n-1}\varphi =(F_{n}+F_{n-1})\varphi +F_{n}=F_{n+1}\varphi +F_{n}.}Forψ=−1/φ{\displaystyle \psi =-1/\varphi }, it is also the case thatψ2=ψ+1{\displaystyle \psi ^{2}=\psi +1} and it is also the case thatψn=Fnψ+Fn−1.{\displaystyle \psi ^{n}=F_{n}\psi +F_{n-1}.}

These expressions are also true forn < 1 if the Fibonacci sequenceFn isextended to negative integers using the Fibonacci ruleFn=Fn+2−Fn+1.{\displaystyle F_{n}=F_{n+2}-F_{n+1}.}

Identification

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Binet's formula provides a proof that a positive integerx is a Fibonacci numberif and only if at least one of5x2+4{\displaystyle 5x^{2}+4} or5x2−4{\displaystyle 5x^{2}-4} is aperfect square.[30] This is because Binet's formula, which can be written asFn=(φn−(−1)nφ−n)/5{\displaystyle F_{n}=(\varphi ^{n}-(-1)^{n}\varphi ^{-n})/{\sqrt {5}}}, can be multiplied by5φn{\displaystyle {\sqrt {5}}\varphi ^{n}} and solved as aquadratic equation inφn{\displaystyle \varphi ^{n}} via thequadratic formula:

φn=Fn5±5Fn2+4(−1)n2.{\displaystyle \varphi ^{n}={\frac {F_{n}{\sqrt {5}}\pm {\sqrt {5{F_{n}}^{2}+4(-1)^{n}}}}{2}}.}

Comparing this toφn=Fnφ+Fn−1=(Fn5+Fn+2Fn−1)/2{\displaystyle \varphi ^{n}=F_{n}\varphi +F_{n-1}=(F_{n}{\sqrt {5}}+F_{n}+2F_{n-1})/2}, it follows that

5Fn2+4(−1)n=(Fn+2Fn−1)2.{\displaystyle 5{F_{n}}^{2}+4(-1)^{n}=(F_{n}+2F_{n-1})^{2}\,.}

In particular, the left-hand side is a perfect square.

Matrix form

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

(Fk+2Fk+1)=(1110)(Fk+1Fk){\displaystyle {F_{k+2} \choose F_{k+1}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{F_{k+1} \choose F_{k}}}alternatively denotedF→k+1=AF→k,{\displaystyle {\vec {F}}_{k+1}=\mathbf {A} {\vec {F}}_{k},}

which yieldsF→n=AnF→0{\displaystyle {\vec {F}}_{n}=\mathbf {A} ^{n}{\vec {F}}_{0}}. Theeigenvalues of thematrixA areφ=12(1+5 ){\displaystyle \varphi ={\tfrac {1}{2}}{\bigl (}1+{\sqrt {5}}~\!{\bigr )}} andψ=−φ−1=12(1−5 ){\displaystyle \psi =-\varphi ^{-1}={\tfrac {1}{2}}{\bigl (}1-{\sqrt {5}}~\!{\bigr )}} corresponding to the respectiveeigenvectorsμ→=(φ1),ν→=(−φ−11).{\displaystyle {\vec {\mu }}={\varphi \choose 1},\quad {\vec {\nu }}={-\varphi ^{-1} \choose 1}.}

As the initial value isF→0=(10)=15μ→−15ν→,{\displaystyle {\vec {F}}_{0}={1 \choose 0}={\frac {1}{\sqrt {5}}}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}{\vec {\nu }},}

it follows that thenth element isF→n=15Anμ→−15Anν→=15φnμ→−15(−φ)−nν→=15(1+52)n(φ1)−15(1−52)n(−φ−11).{\displaystyle {\begin{aligned}{\vec {F}}_{n}&={\frac {1}{\sqrt {5}}}A^{n}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}A^{n}{\vec {\nu }}\\&={\frac {1}{\sqrt {5}}}\varphi ^{n}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}(-\varphi )^{-n}{\vec {\nu }}\\&={\cfrac {1}{\sqrt {5}}}\left({\cfrac {1+{\sqrt {5}}}{2}}\right)^{\!n}{\varphi \choose 1}\,-\,{\cfrac {1}{\sqrt {5}}}\left({\cfrac {1-{\sqrt {5}}}{2}}\right)^{\!n}{-\varphi ^{-1} \choose 1}.\end{aligned}}}

From this, thenth element in the Fibonacci series may be read off directly as aclosed-form expression:Fn=15(1+52)n−15(1−52)n.{\displaystyle F_{n}={\cfrac {1}{\sqrt {5}}}\left({\cfrac {1+{\sqrt {5}}}{2}}\right)^{\!n}-\,{\cfrac {1}{\sqrt {5}}}\left({\cfrac {1-{\sqrt {5}}}{2}}\right)^{\!n}.}

Equivalently, the same computation may be performed bydiagonalization ofA through use of itseigendecomposition:

A=SΛS−1,An=SΛnS−1,{\displaystyle {\begin{aligned}A&=S\Lambda S^{-1},\\[3mu]A^{n}&=S\Lambda ^{n}S^{-1},\end{aligned}}}

where

Λ=(φ00−φ−1),S=(φ−φ−111).{\displaystyle \Lambda ={\begin{pmatrix}\varphi &0\\0&-\varphi ^{-1}\!\end{pmatrix}},\quad S={\begin{pmatrix}\varphi &-\varphi ^{-1}\\1&1\end{pmatrix}}.}

The closed-form expression for thenth element in the Fibonacci series is therefore given by

(Fn+1Fn)=An(F1F0)=SΛnS−1(F1F0)=S(φn00(−φ)−n)S−1(F1F0)=(φ−φ−111)(φn00(−φ)−n)15(1φ−1−1φ)(10),{\displaystyle {\begin{aligned}{F_{n+1} \choose F_{n}}&=A^{n}{F_{1} \choose F_{0}}\\&=S\Lambda ^{n}S^{-1}{F_{1} \choose F_{0}}\\&=S{\begin{pmatrix}\varphi ^{n}&0\\0&(-\varphi )^{-n}\end{pmatrix}}S^{-1}{F_{1} \choose F_{0}}\\&={\begin{pmatrix}\varphi &-\varphi ^{-1}\\1&1\end{pmatrix}}{\begin{pmatrix}\varphi ^{n}&0\\0&(-\varphi )^{-n}\end{pmatrix}}{\frac {1}{\sqrt {5}}}{\begin{pmatrix}1&\varphi ^{-1}\\-1&\varphi \end{pmatrix}}{1 \choose 0},\end{aligned}}}

which again yieldsFn=φn−(−φ)−n5.{\displaystyle F_{n}={\cfrac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}.}

The matrixA has adeterminant of −1, and thus it is a 2 × 2unimodular matrix.

This property can be understood in terms of thecontinued fraction representation for the golden ratioφ:

φ=1+11+11+11+⋱.{\displaystyle \varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}}}.}

Theconvergents of the continued fraction forφ are ratios of successive Fibonacci numbers:φn =Fn+1 /Fn is then-th convergent, and the(n + 1)-st convergent can be found from the recurrence relationφn+1 = 1 + 1 /φn.[31] The matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1. The matrix representation gives the following closed-form expression for the Fibonacci numbers:

(1110)n=(Fn+1FnFnFn−1).{\displaystyle {\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n}={\begin{pmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{pmatrix}}.}

For a givenn, this matrix can be computed inO(logn) arithmetic operations, using theexponentiation by squaring method.

Taking the determinant of both sides of this equation yieldsCassini's identity,(−1)n=Fn+1Fn−1−Fn2.{\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-{F_{n}}^{2}.}

Moreover, sinceAnAm =An+m for anysquare matrixA, the followingidentities can be derived (they are obtained from two different coefficients of thematrix product, and one may easily deduce the second one from the first one by changingn inton + 1),FmFn+Fm−1Fn−1=Fm+n−1,FmFn+1+Fm−1Fn=Fm+n.{\displaystyle {\begin{aligned}{F_{m}}{F_{n}}+{F_{m-1}}{F_{n-1}}&=F_{m+n-1},\\[3mu]F_{m}F_{n+1}+F_{m-1}F_{n}&=F_{m+n}.\end{aligned}}}

In particular, withm =n,F2n−1=Fn2+Fn−12F2n−1=(Fn−1+Fn+1)Fn=(2Fn−1+Fn)Fn=(2Fn+1−Fn)Fn.{\displaystyle {\begin{aligned}F_{2n-1}&={F_{n}}^{2}+{F_{n-1}}^{2}\\[6mu]F_{2n{\phantom {{}-1}}}&=(F_{n-1}+F_{n+1})F_{n}\\[3mu]&=(2F_{n-1}+F_{n})F_{n}\\[3mu]&=(2F_{n+1}-F_{n})F_{n}.\end{aligned}}}

These last two identities provide a way to compute Fibonacci numbersrecursively inO(logn) arithmetic operations. This matches the time for computing then-th Fibonacci number from the closed-form matrix formula, but with fewer redundant steps if one avoids recomputing an already computed Fibonacci number (recursion withmemoization).[32]

Combinatorial identities

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Combinatorial proofs

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Most identities involving Fibonacci numbers can be proved usingcombinatorial arguments using the fact thatFn{\displaystyle F_{n}} can be interpreted as the number of (possibly empty) sequences of 1s and 2s whose sum isn−1{\displaystyle n-1}. This can be taken as the definition ofFn{\displaystyle F_{n}} with the conventionsF0=0{\displaystyle F_{0}=0}, meaning no such sequence exists whose sum is −1, andF1=1{\displaystyle F_{1}=1}, meaning the empty sequence "adds up" to 0. In the following,|...|{\displaystyle |{...}|} is thecardinality of aset:

F0=0=|{}|{\displaystyle F_{0}=0=|\{\}|}
F1=1=|{()}|{\displaystyle F_{1}=1=|\{()\}|}
F2=1=|{(1)}|{\displaystyle F_{2}=1=|\{(1)\}|}
F3=2=|{(1,1),(2)}|{\displaystyle F_{3}=2=|\{(1,1),(2)\}|}
F4=3=|{(1,1,1),(1,2),(2,1)}|{\displaystyle F_{4}=3=|\{(1,1,1),(1,2),(2,1)\}|}
F5=5=|{(1,1,1,1),(1,1,2),(1,2,1),(2,1,1),(2,2)}|{\displaystyle F_{5}=5=|\{(1,1,1,1),(1,1,2),(1,2,1),(2,1,1),(2,2)\}|}

In this manner the recurrence relationFn=Fn−1+Fn−2{\displaystyle F_{n}=F_{n-1}+F_{n-2}}may be understood by dividing theFn{\displaystyle F_{n}} sequences into two non-overlapping sets where all sequences either begin with 1 or 2:Fn=|{(1,...),(1,...),...}|+|{(2,...),(2,...),...}|{\displaystyle F_{n}=|\{(1,...),(1,...),...\}|+|\{(2,...),(2,...),...\}|}Excluding the first element, the remaining terms in each sequence sum ton−2{\displaystyle n-2} orn−3{\displaystyle n-3} and the cardinality of each set isFn−1{\displaystyle F_{n-1}} orFn−2{\displaystyle F_{n-2}} giving a total ofFn−1+Fn−2{\displaystyle F_{n-1}+F_{n-2}} sequences, showing this is equal toFn{\displaystyle F_{n}}.

In a similar manner it may be shown that the sum of the first Fibonacci numbers up to then-th is equal to the(n + 2)-th Fibonacci number minus 1.[33] In symbols:∑i=1nFi=Fn+2−1{\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1}

This may be seen by dividing all sequences summing ton+1{\displaystyle n+1} based on the location of the first 2. Specifically, each set consists of those sequences that start(2,...),(1,2,...),...,{\displaystyle (2,...),(1,2,...),...,} until the last two sets{(1,1,...,1,2)},{(1,1,...,1)}{\displaystyle \{(1,1,...,1,2)\},\{(1,1,...,1)\}} each with cardinality 1.

Following the same logic as before, by summing the cardinality of each set we see that

Fn+2=Fn+Fn−1+...+|{(1,1,...,1,2)}|+|{(1,1,...,1)}|{\displaystyle F_{n+2}=F_{n}+F_{n-1}+...+|\{(1,1,...,1,2)\}|+|\{(1,1,...,1)\}|}

... where the last two terms have the valueF1=1{\displaystyle F_{1}=1}. From this it follows that∑i=1nFi=Fn+2−1{\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1}.

A similar argument, grouping the sums by the position of the first 1 rather than the first 2 gives two more identities:∑i=0n−1F2i+1=F2n{\displaystyle \sum _{i=0}^{n-1}F_{2i+1}=F_{2n}}and∑i=1nF2i=F2n+1−1.{\displaystyle \sum _{i=1}^{n}F_{2i}=F_{2n+1}-1.}In words, the sum of the first Fibonacci numbers withodd index up toF2n−1{\displaystyle F_{2n-1}} is the(2n)-th Fibonacci number, and the sum of the first Fibonacci numbers witheven index up toF2n{\displaystyle F_{2n}} is the(2n + 1)-th Fibonacci number minus 1.[34]

A different trick may be used to prove∑i=1nFi2=FnFn+1{\displaystyle \sum _{i=1}^{n}F_{i}^{2}=F_{n}F_{n+1}}or in words, the sum of the squares of the first Fibonacci numbers up toFn{\displaystyle F_{n}} is the product of then-th and(n + 1)-th Fibonacci numbers. To see this, begin with a Fibonacci rectangle of sizeFn×Fn+1{\displaystyle F_{n}\times F_{n+1}} and decompose it into squares of sizeFn,Fn−1,...,F1{\displaystyle F_{n},F_{n-1},...,F_{1}}; from this the identity follows by comparingareas:

Symbolic method

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The sequence(Fn)n∈N{\displaystyle (F_{n})_{n\in \mathbb {N} }} is also considered using thesymbolic method.[35] More precisely, this sequence corresponds to aspecifiable combinatorial class. The specification of this sequence isSeq⁡(Z+Z2){\displaystyle \operatorname {Seq} ({\mathcal {Z+Z^{2}}})}. Indeed, as stated above, then{\displaystyle n}-th Fibonacci number equals the number ofcombinatorial compositions (orderedpartitions) ofn−1{\displaystyle n-1} using terms 1 and 2.

It follows that theordinary generating function of the Fibonacci sequence,∑i=0∞Fizi{\displaystyle \sum _{i=0}^{\infty }F_{i}z^{i}}, is therational functionz1−z−z2.{\displaystyle {\frac {z}{1-z-z^{2}}}.}

Induction proofs

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Fibonacci identities often can be easily proved usingmathematical induction.

For example, reconsider∑i=1nFi=Fn+2−1.{\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1.}AddingFn+1{\displaystyle F_{n+1}} to both sides gives

∑i=1nFi+Fn+1=Fn+1+Fn+2−1{\displaystyle \sum _{i=1}^{n}F_{i}+F_{n+1}=F_{n+1}+F_{n+2}-1}

and so we have the formula forn+1{\displaystyle n+1}∑i=1n+1Fi=Fn+3−1{\displaystyle \sum _{i=1}^{n+1}F_{i}=F_{n+3}-1}

Similarly, addFn+12{\displaystyle {F_{n+1}}^{2}} to both sides of∑i=1nFi2=FnFn+1{\displaystyle \sum _{i=1}^{n}F_{i}^{2}=F_{n}F_{n+1}}to give∑i=1nFi2+Fn+12=Fn+1(Fn+Fn+1){\displaystyle \sum _{i=1}^{n}F_{i}^{2}+{F_{n+1}}^{2}=F_{n+1}\left(F_{n}+F_{n+1}\right)}∑i=1n+1Fi2=Fn+1Fn+2{\displaystyle \sum _{i=1}^{n+1}F_{i}^{2}=F_{n+1}F_{n+2}}

Binet formula proofs

[edit]

The Binet formula is5Fn=φn−ψn.{\displaystyle {\sqrt {5}}F_{n}=\varphi ^{n}-\psi ^{n}.}This can be used to prove Fibonacci identities.

For example, to prove that∑i=1nFi=Fn+2−1{\textstyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1}note that the left hand side multiplied by5{\displaystyle {\sqrt {5}}} becomes1+φ+φ2+⋯+φn−(1+ψ+ψ2+⋯+ψn)=φn+1−1φ−1−ψn+1−1ψ−1=φn+1−1−ψ−ψn+1−1−φ=−φn+2+φ+ψn+2−ψφψ=φn+2−ψn+2−(φ−ψ)=5(Fn+2−1){\displaystyle {\begin{aligned}1+&\varphi +\varphi ^{2}+\dots +\varphi ^{n}-\left(1+\psi +\psi ^{2}+\dots +\psi ^{n}\right)\\&={\frac {\varphi ^{n+1}-1}{\varphi -1}}-{\frac {\psi ^{n+1}-1}{\psi -1}}\\&={\frac {\varphi ^{n+1}-1}{-\psi }}-{\frac {\psi ^{n+1}-1}{-\varphi }}\\&={\frac {-\varphi ^{n+2}+\varphi +\psi ^{n+2}-\psi }{\varphi \psi }}\\&=\varphi ^{n+2}-\psi ^{n+2}-(\varphi -\psi )\\&={\sqrt {5}}(F_{n+2}-1)\\\end{aligned}}}as required, using the factsφψ=−1{\textstyle \varphi \psi =-1} andφ−ψ=5{\textstyle \varphi -\psi ={\sqrt {5}}} to simplify the equations.

Other identities

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Numerous other identities can be derived using various methods. Here are some of them:[36]

Cassini's and Catalan's identities

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Main article:Cassini and Catalan identities

Cassini's identity states thatFn2−Fn+1Fn−1=(−1)n−1{\displaystyle {F_{n}}^{2}-F_{n+1}F_{n-1}=(-1)^{n-1}}Catalan's identity is a generalization:Fn2−Fn+rFn−r=(−1)n−rFr2{\displaystyle {F_{n}}^{2}-F_{n+r}F_{n-r}=(-1)^{n-r}{F_{r}}^{2}}

d'Ocagne's identity

[edit]

FmFn+1−Fm+1Fn=(−1)nFm−n{\displaystyle F_{m}F_{n+1}-F_{m+1}F_{n}=(-1)^{n}F_{m-n}}F2n=Fn+12−Fn−12=Fn(Fn+1+Fn−1)=FnLn{\displaystyle F_{2n}={F_{n+1}}^{2}-{F_{n-1}}^{2}=F_{n}\left(F_{n+1}+F_{n-1}\right)=F_{n}L_{n}}whereLn is then-thLucas number. The last is an identity for doublingn; other identities of this type areF3n=2Fn3+3FnFn+1Fn−1=5Fn3+3(−1)nFn{\displaystyle F_{3n}=2{F_{n}}^{3}+3F_{n}F_{n+1}F_{n-1}=5{F_{n}}^{3}+3(-1)^{n}F_{n}}by Cassini's identity.

F3n+1=Fn+13+3Fn+1Fn2−Fn3{\displaystyle F_{3n+1}={F_{n+1}}^{3}+3F_{n+1}{F_{n}}^{2}-{F_{n}}^{3}}F3n+2=Fn+13+3Fn+12Fn+Fn3{\displaystyle F_{3n+2}={F_{n+1}}^{3}+3{F_{n+1}}^{2}F_{n}+{F_{n}}^{3}}F4n=4FnFn+1(Fn+12+2Fn2)−3Fn2(Fn2+2Fn+12){\displaystyle F_{4n}=4F_{n}F_{n+1}\left({F_{n+1}}^{2}+2{F_{n}}^{2}\right)-3{F_{n}}^{2}\left({F_{n}}^{2}+2{F_{n+1}}^{2}\right)}These can be found experimentally usinglattice reduction, and are useful in setting up thespecial number field sieve tofactorize a Fibonacci number.

More generally,[36]

Fkn+c=∑i=0k(ki)Fc−iFniFn+1k−i.{\displaystyle F_{kn+c}=\sum _{i=0}^{k}{k \choose i}F_{c-i}{F_{n}}^{i}{F_{n+1}}^{k-i}.}

or alternatively

Fkn+c=∑i=0k(ki)Fc+iFniFn−1k−i.{\displaystyle F_{kn+c}=\sum _{i=0}^{k}{k \choose i}F_{c+i}{F_{n}}^{i}{F_{n-1}}^{k-i}.}

Puttingk = 2 in this formula, one gets again the formulas of the end of above sectionMatrix form.

Generating function

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Thegenerating function of the Fibonacci sequence is thepower series

s(z)=∑k=0∞Fkzk=0+z+z2+2z3+3z4+5z5+….{\displaystyle s(z)=\sum _{k=0}^{\infty }F_{k}z^{k}=0+z+z^{2}+2z^{3}+3z^{4}+5z^{5}+\dots .}

This series is convergent for anycomplex numberz{\displaystyle z} satisfying|z|<1/φ≈0.618,{\displaystyle |z|<1/\varphi \approx 0.618,} and its sum has a simple closed form:[37]

s(z)=z1−z−z2.{\displaystyle s(z)={\frac {z}{1-z-z^{2}}}.}

This can be proved by multiplying by(1−z−z2){\textstyle (1-z-z^{2})}:(1−z−z2)s(z)=∑k=0∞Fkzk−∑k=0∞Fkzk+1−∑k=0∞Fkzk+2=∑k=0∞Fkzk−∑k=1∞Fk−1zk−∑k=2∞Fk−2zk=0z0+1z1−0z1+∑k=2∞(Fk−Fk−1−Fk−2)zk=z,{\displaystyle {\begin{aligned}(1-z-z^{2})s(z)&=\sum _{k=0}^{\infty }F_{k}z^{k}-\sum _{k=0}^{\infty }F_{k}z^{k+1}-\sum _{k=0}^{\infty }F_{k}z^{k+2}\\&=\sum _{k=0}^{\infty }F_{k}z^{k}-\sum _{k=1}^{\infty }F_{k-1}z^{k}-\sum _{k=2}^{\infty }F_{k-2}z^{k}\\&=0z^{0}+1z^{1}-0z^{1}+\sum _{k=2}^{\infty }(F_{k}-F_{k-1}-F_{k-2})z^{k}\\&=z,\end{aligned}}}

where all terms involvingzk{\displaystyle z^{k}} fork≥2{\displaystyle k\geq 2} cancel out because of the defining Fibonacci recurrence relation.

Thepartial fraction decomposition is given bys(z)=15(11−φz−11−ψz){\displaystyle s(z)={\frac {1}{\sqrt {5}}}\left({\frac {1}{1-\varphi z}}-{\frac {1}{1-\psi z}}\right)}whereφ=12(1+5){\textstyle \varphi ={\tfrac {1}{2}}\left(1+{\sqrt {5}}\right)} is the golden ratio andψ=12(1−5){\displaystyle \psi ={\tfrac {1}{2}}\left(1-{\sqrt {5}}\right)} is itsconjugate.

The related functionz↦−s(−1/z){\textstyle z\mapsto -s\left(-1/z\right)} is the generating function for thenegafibonacci numbers, ands(z){\displaystyle s(z)} satisfies thefunctional equation

s(z)=s(−1z).{\displaystyle s(z)=s\!\left(-{\frac {1}{z}}\right).}

Usingz{\displaystyle z} equal to any of 0.01, 0.001, 0.0001, etc. lays out the first Fibonacci numbers in the decimal expansion ofs(z){\displaystyle s(z)}. For example,s(0.001)=0.0010.998999=1000998999=0.001001002003005008013021….{\displaystyle s(0.001)={\frac {0.001}{0.998999}}={\frac {1000}{998999}}=0.001001002003005008013021\ldots .}

Reciprocal sums

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Infinite sums overreciprocal Fibonacci numbers can sometimes be evaluated in terms oftheta functions. For example, the sum of every odd-indexed reciprocal Fibonacci number can be written as∑k=1∞1F2k−1=54ϑ2(0,3−52)2,{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{2k-1}}}={\frac {\sqrt {5}}{4}}\;\vartheta _{2}\!\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)^{2},}

and the sum of squared reciprocal Fibonacci numbers as∑k=1∞1Fk2=524(ϑ2(0,3−52)4−ϑ4(0,3−52)4+1).{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{{F_{k}}^{2}}}={\frac {5}{24}}\!\left(\vartheta _{2}\!\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)^{4}-\vartheta _{4}\!\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)^{4}+1\right).}

If we add 1 to each Fibonacci number in the first sum, there is also the closed form∑k=1∞11+F2k−1=52,{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{1+F_{2k-1}}}={\frac {\sqrt {5}}{2}},}

and there is anested sum of squared Fibonacci numbers giving the reciprocal of thegolden ratio,∑k=1∞(−1)k+1∑j=1kFj2=5−12.{\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{\sum _{j=1}^{k}{F_{j}}^{2}}}={\frac {{\sqrt {5}}-1}{2}}.}

The sum of all even-indexed reciprocal Fibonacci numbers is[38]∑k=1∞1F2k=5(L(ψ2)−L(ψ4)){\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{2k}}}={\sqrt {5}}\left(L(\psi ^{2})-L(\psi ^{4})\right)}with theLambert seriesL(q):=∑k=1∞qk1−qk,{\displaystyle \textstyle L(q):=\sum _{k=1}^{\infty }{\frac {q^{k}}{1-q^{k}}},} since1F2k=5(ψ2k1−ψ2k−ψ4k1−ψ4k).{\displaystyle \textstyle {\frac {1}{F_{2k}}}={\sqrt {5}}\left({\frac {\psi ^{2k}}{1-\psi ^{2k}}}-{\frac {\psi ^{4k}}{1-\psi ^{4k}}}\right)\!.}

So thereciprocal Fibonacci constant is[39]∑k=1∞1Fk=∑k=1∞1F2k−1+∑k=1∞1F2k=3.359885666243…{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{k}}}=\sum _{k=1}^{\infty }{\frac {1}{F_{2k-1}}}+\sum _{k=1}^{\infty }{\frac {1}{F_{2k}}}=3.359885666243\dots }

Moreover, this number has been provedirrational byRichard André-Jeannin.[40]

Millin's series gives the identity[41]∑k=0∞1F2k=7−52,{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{F_{2^{k}}}}={\frac {7-{\sqrt {5}}}{2}},}which follows from the closed form for its partial sums asN tends to infinity:∑k=0N1F2k=3−F2N−1F2N.{\displaystyle \sum _{k=0}^{N}{\frac {1}{F_{2^{k}}}}=3-{\frac {F_{2^{N}-1}}{F_{2^{N}}}}.}

Primes and divisibility

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Divisibility properties

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Every third number of the sequence is even (a multiple ofF3=2{\displaystyle F_{3}=2}) and, more generally, everyk-th number of the sequence is a multiple ofFk. Thus the Fibonacci sequence is an example of adivisibility sequence. In fact, the Fibonacci sequence satisfies the stronger divisibility property[42][43]gcd(Fa,Fb,Fc,…)=Fgcd(a,b,c,…){\displaystyle \gcd(F_{a},F_{b},F_{c},\ldots )=F_{\gcd(a,b,c,\ldots )}\,}wheregcd is thegreatest common divisor function. (This relation is different if a different indexing convention is used, such as the one that starts the sequence with⁠F0=1{\displaystyle F_{0}=1}⁠ and⁠F1=1{\displaystyle F_{1}=1}⁠.)

In particular, any three consecutive Fibonacci numbers are pairwisecoprime because bothF1=1{\displaystyle F_{1}=1} andF2=1{\displaystyle F_{2}=1}. That is,

gcd(Fn,Fn+1)=gcd(Fn,Fn+2)=gcd(Fn+1,Fn+2)=1{\displaystyle \gcd(F_{n},F_{n+1})=\gcd(F_{n},F_{n+2})=\gcd(F_{n+1},F_{n+2})=1}

for everyn.

Everyprime numberp divides a Fibonacci number that can be determined by the value ofpmodulo 5. Ifp is congruent to 1 or 4 modulo 5, thenp dividesFp−1, and ifp is congruent to 2 or 3 modulo 5, then,p dividesFp+1. The remaining case is thatp = 5, and in this casep dividesFp.

{p=5⇒p∣Fp,p≡±1(mod5)⇒p∣Fp−1,p≡±2(mod5)⇒p∣Fp+1.{\displaystyle {\begin{cases}p=5&\Rightarrow p\mid F_{p},\\p\equiv \pm 1{\pmod {5}}&\Rightarrow p\mid F_{p-1},\\p\equiv \pm 2{\pmod {5}}&\Rightarrow p\mid F_{p+1}.\end{cases}}}

These cases can be combined into a single, non-piecewise formula, using theLegendre symbol:[44]p∣Fp−(5p).{\displaystyle p\mid F_{p\;-\,\left({\frac {5}{p}}\right)}.}

Primality testing

[edit]

The above formula can be used as aprimality test in the sense that ifn∣Fn−(5n),{\displaystyle n\mid F_{n\;-\,\left({\frac {5}{n}}\right)},}where the Legendre symbol has been replaced by theJacobi symbol, then this is evidence thatn is a prime, and if it fails to hold, thenn is definitely not a prime. Ifn iscomposite and satisfies the formula, thenn is aFibonacci pseudoprime. Whenm is large – say a 500-bit number – then we can calculateFm (modn) efficiently using the matrix form. Thus

(Fm+1FmFmFm−1)≡(1110)m(modn).{\displaystyle {\begin{pmatrix}F_{m+1}&F_{m}\\F_{m}&F_{m-1}\end{pmatrix}}\equiv {\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{m}{\pmod {n}}.}Here the matrix powerAm is calculated usingmodular exponentiation, which can beadapted to matrices.[45]

Fibonacci primes

[edit]
Main article:Fibonacci prime

AFibonacci prime is a Fibonacci number that isprime. The first few are:[46]

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

Fibonacci primes with thousands of digits have been found, but it is not known whether there are infinitely many.[47]

Fkn is divisible byFn, so, apart fromF4 = 3, any Fibonacci prime must have a prime index. As there arearbitrarily long runs ofcomposite numbers, there are therefore also arbitrarily long runs of composite Fibonacci numbers.

No Fibonacci number greater thanF6 = 8 is one greater or one less than a prime number.[48]

The only nontrivialsquare Fibonacci number is 144.[49] Attila Pethő proved in 2001 that there is only a finite number ofperfect power Fibonacci numbers.[50] In 2006, Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only such non-trivial perfect powers.[51]

1, 3, 21, and 55 are the onlytriangular Fibonacci numbers, which wasconjectured byVern Hoggatt and proved by Luo Ming.[52]

No Fibonacci number can be aperfect number.[53] More generally, no Fibonacci number other than 1 can bemultiply perfect,[54] and no ratio of two Fibonacci numbers can be perfect.[55]

Prime divisors

[edit]

With the exceptions of 1, 8 and 144 (F1 =F2,F6 andF12) every Fibonacci number has a prime factor that is not a factor of any smaller Fibonacci number (Carmichael's theorem).[56] As a result, 8 and 144 (F6 andF12) are the only Fibonacci numbers that are the product of other Fibonacci numbers.[57]

The divisibility of Fibonacci numbers by a primep is related to theLegendre symbol(p5){\displaystyle {\bigl (}{\tfrac {p}{5}}{\bigr )}} which is evaluated as follows:(p5)={0if p=51if p≡±1(mod5)−1if p≡±2(mod5).{\displaystyle \left({\frac {p}{5}}\right)={\begin{cases}0&{\text{if }}p=5\\1&{\text{if }}p\equiv \pm 1{\pmod {5}}\\-1&{\text{if }}p\equiv \pm 2{\pmod {5}}.\end{cases}}}

Ifp is a prime number thenFp≡(p5)(modp)andFp−(p5)≡0(modp).{\displaystyle F_{p}\equiv \left({\frac {p}{5}}\right){\pmod {p}}\quad {\text{and}}\quad F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p}}.}[58][59]

For example,(25)=−1,F3=2,F2=1,(35)=−1,F4=3,F3=2,(55)=0,F5=5,(75)=−1,F8=21,F7=13,(115)=+1,F10=55,F11=89.{\displaystyle {\begin{aligned}{\bigl (}{\tfrac {2}{5}}{\bigr )}&=-1,&F_{3}&=2,&F_{2}&=1,\\{\bigl (}{\tfrac {3}{5}}{\bigr )}&=-1,&F_{4}&=3,&F_{3}&=2,\\{\bigl (}{\tfrac {5}{5}}{\bigr )}&=0,&F_{5}&=5,\\{\bigl (}{\tfrac {7}{5}}{\bigr )}&=-1,&F_{8}&=21,&F_{7}&=13,\\{\bigl (}{\tfrac {11}{5}}{\bigr )}&=+1,&F_{10}&=55,&F_{11}&=89.\end{aligned}}}

It is not known whether there exists a primep such that

Fp−(p5)≡0(modp2).{\displaystyle F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p^{2}}}.}

Such primes (if there are any) would be calledWall–Sun–Sun primes.

Also, ifp ≠ 5 is an odd prime number then:[60]5Fp±122≡{12(5(p5)±5)(modp)if p≡1(mod4)12(5(p5)∓3)(modp)if p≡3(mod4).{\displaystyle 5{F_{\frac {p\pm 1}{2}}}^{2}\equiv {\begin{cases}{\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {p}{5}}{\bigr )}\pm 5\right){\pmod {p}}&{\text{if }}p\equiv 1{\pmod {4}}\\{\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {p}{5}}{\bigr )}\mp 3\right){\pmod {p}}&{\text{if }}p\equiv 3{\pmod {4}}.\end{cases}}}

Example 1.p = 7, in this casep ≡ 3 (mod 4) and we have:(75)=−1:12(5(75)+3)=−1,12(5(75)−3)=−4.{\displaystyle {\bigl (}{\tfrac {7}{5}}{\bigr )}=-1:\qquad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {7}{5}}{\bigr )}+3\right)=-1,\quad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {7}{5}}{\bigr )}-3\right)=-4.}F3=2 and F4=3.{\displaystyle F_{3}=2{\text{ and }}F_{4}=3.}5F32=20≡−1(mod7) and 5F42=45≡−4(mod7){\displaystyle 5{F_{3}}^{2}=20\equiv -1{\pmod {7}}\;\;{\text{ and }}\;\;5{F_{4}}^{2}=45\equiv -4{\pmod {7}}}

Example 2.p = 11, in this casep ≡ 3 (mod 4) and we have:(115)=+1:12(5(115)+3)=4,12(5(115)−3)=1.{\displaystyle {\bigl (}{\tfrac {11}{5}}{\bigr )}=+1:\qquad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {11}{5}}{\bigr )}+3\right)=4,\quad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {11}{5}}{\bigr )}-3\right)=1.}F5=5 and F6=8.{\displaystyle F_{5}=5{\text{ and }}F_{6}=8.}5F52=125≡4(mod11) and 5F62=320≡1(mod11){\displaystyle 5{F_{5}}^{2}=125\equiv 4{\pmod {11}}\;\;{\text{ and }}\;\;5{F_{6}}^{2}=320\equiv 1{\pmod {11}}}

Example 3.p = 13, in this casep ≡ 1 (mod 4) and we have:(135)=−1:12(5(135)−5)=−5,12(5(135)+5)=0.{\displaystyle {\bigl (}{\tfrac {13}{5}}{\bigr )}=-1:\qquad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {13}{5}}{\bigr )}-5\right)=-5,\quad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {13}{5}}{\bigr )}+5\right)=0.}F6=8 and F7=13.{\displaystyle F_{6}=8{\text{ and }}F_{7}=13.}5F62=320≡−5(mod13) and 5F72=845≡0(mod13){\displaystyle 5{F_{6}}^{2}=320\equiv -5{\pmod {13}}\;\;{\text{ and }}\;\;5{F_{7}}^{2}=845\equiv 0{\pmod {13}}}

Example 4.p = 29, in this casep ≡ 1 (mod 4) and we have:(295)=+1:12(5(295)−5)=0,12(5(295)+5)=5.{\displaystyle {\bigl (}{\tfrac {29}{5}}{\bigr )}=+1:\qquad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {29}{5}}{\bigr )}-5\right)=0,\quad {\tfrac {1}{2}}\left(5{\bigl (}{\tfrac {29}{5}}{\bigr )}+5\right)=5.}F14=377 and F15=610.{\displaystyle F_{14}=377{\text{ and }}F_{15}=610.}5F142=710645≡0(mod29) and 5F152=1860500≡5(mod29){\displaystyle 5{F_{14}}^{2}=710645\equiv 0{\pmod {29}}\;\;{\text{ and }}\;\;5{F_{15}}^{2}=1860500\equiv 5{\pmod {29}}}

For oddn, all odd prime divisors ofFn are congruent to 1 modulo 4, implying that all odd divisors ofFn (as the products of odd prime divisors) are congruent to 1 modulo 4.[61]

For example,F1=1, F3=2, F5=5, F7=13, F9=34=2⋅17, F11=89, F13=233, F15=610=2⋅5⋅61.{\displaystyle F_{1}=1,\ F_{3}=2,\ F_{5}=5,\ F_{7}=13,\ F_{9}={\color {Red}34}=2\cdot 17,\ F_{11}=89,\ F_{13}=233,\ F_{15}={\color {Red}610}=2\cdot 5\cdot 61.}

All known factors of Fibonacci numbersF(i) for alli < 50000 are collected at the relevant repositories.[62][63]

Periodicity modulon

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Main article:Pisano period

If the members of the Fibonacci sequence are taken mod n, the resulting sequence isperiodic with period at most 6n.[64] The lengths of the periods for variousn form the so-calledPisano periods.[65] Determining a general formula for the Pisano periods is anopen problem, which includes as a subproblem a special instance of the problem of finding themultiplicative order of amodular integer or of an element in afinite field. However, for any particularn, the Pisano period may be found as an instance ofcycle detection.

Generalizations

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Main article:Generalizations of Fibonacci numbers

The Fibonacci sequence is one of the simplest and earliest known sequences defined by arecurrence relation, and specifically by a lineardifference equation. All these sequences may be viewed as generalizations of the Fibonacci sequence. In particular, Binet's formula may be generalized to any sequence that is a solution of ahomogeneous linear difference equation with constant coefficients.

Some specific examples that are close, in some sense, to the Fibonacci sequence include:

  • Generalizing the index to negative integers to produce thenegafibonacci numbers.
  • Generalizing the index toreal numbers using a modification of Binet's formula.[36]
  • Starting with other integers.Lucas numbers haveL1 = 1,L2 = 3, andLn =Ln−1 +Ln−2.Primefree sequences use the Fibonacci recursion with other starting points 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. ThePell numbers havePn = 2Pn−1 +Pn−2. If the coefficient of the preceding value is assigned a variable valuex, the result is the sequence ofFibonacci polynomials.
  • Not adding the immediately preceding numbers. ThePadovan sequence andPerrin numbers haveP(n) =P(n − 2) +P(n − 3).
  • Generating the next number by adding 3 numbers (tribonacci numbers), 4 numbers (tetranacci numbers), or more. The resulting sequences are known asn-Step Fibonacci numbers.[66]

Applications

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Mathematics

[edit]
The Fibonacci numbers are the sums of the diagonals (shown in red) of a left-justifiedPascal's triangle.

The Fibonacci numbers occur as the sums ofbinomial coefficients in the "shallow" diagonals ofPascal's triangle:[67]Fn=∑k=0⌊n−12⌋(n−k−1k).{\displaystyle F_{n}=\sum _{k=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }{\binom {n-k-1}{k}}.}This can be proved by expanding the generating functionx1−x−x2=x+x2(1+x)+x3(1+x)2+⋯+xk+1(1+x)k+⋯=∑n=0∞Fnxn{\displaystyle {\frac {x}{1-x-x^{2}}}=x+x^{2}(1+x)+x^{3}(1+x)^{2}+\dots +x^{k+1}(1+x)^{k}+\dots =\sum \limits _{n=0}^{\infty }F_{n}x^{n}}and collecting like terms ofxn{\displaystyle x^{n}}.

To see how the formula is used, we can arrange the sums by the number of terms present:

5= 1+1+1+1+1
= 2+1+1+1= 1+2+1+1= 1+1+2+1= 1+1+1+2
= 2+2+1= 2+1+2= 1+2+2

which is(50)+(41)+(32){\displaystyle \textstyle {\binom {5}{0}}+{\binom {4}{1}}+{\binom {3}{2}}}, where we are choosing the positions ofk twos fromn−k−1 terms.

Use of the Fibonacci sequence to count{1, 2}-restricted compositions

These numbers also give the solution to certain enumerative problems,[68] the most common of which is that of counting the number of ways of writing a given numbern as an ordered sum of 1s and 2s (calledcompositions); there areFn+1 ways to do this (equivalently, it's also the number ofdomino tilings of the2×n{\displaystyle 2\times n} rectangle). For example, there areF5+1 =F6 = 8 ways one can climb a staircase of 5 steps, taking one or two steps at a time:

5= 1+1+1+1+1= 2+1+1+1= 1+2+1+1= 1+1+2+1= 2+2+1
= 1+1+1+2= 2+1+2= 1+2+2

The figure shows that 8 can be decomposed into 5 (the number of ways to climb 4 steps, followed by a single-step) plus 3 (the number of ways to climb 3 steps, followed by a double-step). The same reasoning is appliedrecursively until a single step, of which there is only one way to climb.

The Fibonacci numbers can be found in different ways among the set ofbinarystrings, or equivalently, among thesubsets of a given set.

  • The number of binary strings of lengthn without consecutive1s is the Fibonacci numberFn+2. For example, out of the 16 binary strings of length 4, there areF6 = 8 without consecutive1s—they are 0000, 0001, 0010, 0100, 0101, 1000, 1001, and 1010. Such strings are the binary representations ofFibbinary numbers. Equivalently,Fn+2 is the number of subsetsS of{1, ...,n} without consecutive integers, that is, thoseS for which{i,i + 1} ⊈S for everyi. Abijection with the sums ton+1 is to replace 1 with 0 and 2 with 10, and drop the last zero.
  • The number of binary strings of lengthn without an odd number of consecutive1s is the Fibonacci numberFn+1. For example, out of the 16 binary strings of length 4, there areF5 = 5 without an odd number of consecutive1s—they are 0000, 0011, 0110, 1100, 1111. Equivalently, the number of subsetsS of{1, ...,n} without an odd number of consecutive integers isFn+1. A bijection with the sums ton is to replace 1 with 0 and 2 with 11.
  • The number of binary strings of lengthn without an even number of consecutive0s or1s is2Fn. For example, out of the 16 binary strings of length 4, there are2F4 = 6 without an even number of consecutive0s or1s—they are 0001, 0111, 0101, 1000, 1010, 1110. There is an equivalent statement about subsets.
  • Yuri Matiyasevich was able to show that the Fibonacci numbers can be defined by aDiophantine equation, which led tohis solvingHilbert's tenth problem.[69]
  • The Fibonacci numbers are also an example of acomplete sequence. This means that every positive integer can be written as a sum of Fibonacci numbers, where any one number is used once at most.
  • Moreover, every positive integer can be written in a unique way as the sum ofone or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. This is known asZeckendorf's theorem, and a sum of Fibonacci numbers that satisfies these conditions is called a Zeckendorf representation. The Zeckendorf representation of a number can be used to derive itsFibonacci coding.
  • Starting with 5, every second Fibonacci number is the length of thehypotenuse of aright triangle with integer sides, or in other words, the largest number in aPythagorean triple, obtained from the formula(FnFn+3)2+(2Fn+1Fn+2)2=F2n+32.{\displaystyle (F_{n}F_{n+3})^{2}+(2F_{n+1}F_{n+2})^{2}={F_{2n+3}}^{2}.} The sequence of Pythagorean triangles obtained from this formula has sides of lengths (3,4,5), (5,12,13), (16,30,34), (39,80,89), ... . The middle side of each of these triangles is the sum of the three sides of the preceding triangle.[70]
  • TheFibonacci cube is anundirected graph with a Fibonacci number of nodes that has been proposed as anetwork topology forparallel computing.
  • Fibonacci numbers appear in thering lemma, used to prove connections between thecircle packing theorem andconformal maps.[71]

Computer science

[edit]
Fibonacci tree of height 6.Balance factors green; heights red.
The keys in the left spine are Fibonacci numbers.

Nature

[edit]
Further information:Patterns in nature
See also:Golden ratio § Nature
Yellow chamomile head showing the arrangement in 21 (blue) and 13 (cyan) spirals. Such arrangements involving consecutive Fibonacci numbers appear in a wide variety of plants.

Fibonacci sequences appear in biological settings,[78] such as branching in trees,arrangement of leaves on a stem, the fruitlets of apineapple,[79] the flowering ofartichoke, the arrangement of apine cone,[80] and the family tree ofhoneybees.[81][82]Kepler pointed out the presence of the Fibonacci sequence in nature, using it to explain the (golden ratio-related)pentagonal form of some flowers.[83] Fielddaisies most often have petals in counts of Fibonacci numbers.[84] In 1830,Karl Friedrich Schimper andAlexander Braun discovered that theparastichies (spiralphyllotaxis) of plants were frequently expressed as fractions involving Fibonacci numbers.[85]

Przemysław Prusinkiewicz advanced the idea that real instances can in part be understood as the expression of certain algebraic constraints onfree groups, specifically as certainLindenmayer grammars.[86]

Illustration of Vogel's model forn = 1 ... 500

A model for the pattern offlorets in the head of asunflower was proposed byHelmut Vogel [de] in 1979.[87] This has the form

θ=2πφ2n, r=cn{\displaystyle \theta ={\frac {2\pi }{\varphi ^{2}}}n,\ r=c{\sqrt {n}}}

wheren is the index number of the floret andc is a constant scaling factor; the florets thus lie onFermat's spiral. The divergenceangle, approximately 137.51°, is thegolden 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 formF( j):F( j + 1), the nearest neighbors of floret numbern are those atn ±F( j) for some indexj, which depends onr, the distance from the center. Sunflowers and similar flowers most commonly have spirals of florets in clockwise and counter-clockwise directions in the amount of adjacent Fibonacci numbers,[88] typically counted by the outermost range of radii.[89]

Fibonacci numbers also appear in the ancestral pedigrees ofbees (which arehaplodiploids), according to the following rules:

  • If an egg is laid but not fertilized, it produces a male (ordrone bee in honeybees).
  • If, however, an egg is fertilized, it produces a female.

Thus, a male bee always has one parent, and a female bee has two. If one traces the pedigree of any male bee (1 bee), he has 1 parent (1 bee), 2 grandparents, 3 great-grandparents, 5 great-great-grandparents, and so on. This sequence of numbers of parents is the Fibonacci sequence. The number of ancestors at each level,Fn, is the number of female ancestors, which isFn−1, plus the number of male ancestors, which isFn−2.[90][91] This is under the unrealistic assumption that the ancestors at each level are otherwise unrelated.

The number of possible ancestors on the X chromosome inheritance line at a given ancestral generation follows the Fibonacci sequence. (After Hutchison, L. "Growing the Family Tree: The Power of DNA in Reconstructing Family Relationships".[92])

It has similarly been noticed that the number of possible ancestors on the humanX chromosome inheritance line at a given ancestral generation also follows the Fibonacci sequence.[92] A male individual has an X chromosome, which he received from his mother, and aY chromosome, which he received from his father. The male counts as the "origin" of his own X chromosome (F1=1{\displaystyle F_{1}=1}), and at his parents' generation, his X chromosome came from a single parent(F2=1{\displaystyle F_{2}=1}). The male's mother received one X chromosome from her mother (the son's maternal grandmother), and one from her father (the son's maternal grandfather), so two grandparents contributed to the male descendant's X chromosome(F3=2{\displaystyle F_{3}=2}). The maternal grandfather received his X chromosome from his mother, and the maternal grandmother received X chromosomes from both of her parents, so three great-grandparents contributed to the male descendant's X chromosome(F4=3{\displaystyle F_{4}=3}). Five great-great-grandparents contributed to the male descendant's X chromosome(F5=5{\displaystyle F_{5}=5}), etc. (This assumes that all ancestors of a given descendant are independent, but if any genealogy is traced far enough back in time, ancestors begin to appear on multiple lines of the genealogy, until eventually apopulation founder appears on all lines of the genealogy.)

The Fibonacci sequence can also be found in man-made construction, as seen when looking at the staircase inside the Berlin Victory Column.

Other

[edit]
  • Inoptics, when a beam of light shines at an angle through two stacked transparent plates of different materials of differentrefractive indexes, it may reflect off three surfaces: the top, middle, and bottom surfaces of the two plates. The number of different beam paths that havek reflections, fork > 1, is thek-th Fibonacci number. (However, whenk = 1, there are three reflection paths, not two, one for each of the three surfaces.)[93]
  • Fibonacci retracement levels are widely used intechnical analysis for financial market trading.
  • Since theconversion factor 1.609344 for miles to kilometers is close to the golden ratio, 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 aradix 2 numberregister ingolden ratio baseφ being shifted. To convert from kilometers to miles, shift the register down the Fibonacci sequence instead.[94]
  • The measured values of voltages and currents in the infinite resistor chain circuit (also called theresistor ladder or infinite series-parallel circuit) follow the Fibonacci sequence. The intermediate results of adding the alternating series and parallel resistances yields fractions composed of consecutive Fibonacci numbers. The equivalent resistance of the entire circuit equals the golden ratio.[95]
  • Brasch et al. 2012 show how a generalized Fibonacci sequence also can be connected to the field ofeconomics.[96] In particular, it is shown how a generalized Fibonacci sequence enters the control function of finite-horizon dynamic optimisation problems with one state and one control variable. The procedure is illustrated in an example often referred to as the Brock–Mirman economic growth model.
  • Mario Merz included the Fibonacci sequence in some of his artworks beginning in 1970.[97]
  • Joseph Schillinger (1895–1943) developeda system of composition which uses Fibonacci intervals in some of its melodies; he viewed these as the musical counterpart to the elaborate harmony evident within nature.[98] See alsoGolden ratio § Music.

See also

[edit]

References

[edit]

Explanatory footnotes

[edit]
  1. ^"For four, variations of meters of two [and] three being mixed, five happens. For five, variations of two earlier—three [and] four, being mixed, eight is obtained. In this way, for six, [variations] of four [and] of five being mixed, thirteen happens. And like that, variations of two earlier meters being mixed, sevenmorae [is] twenty-one. In this way, the process should be followed in all mātrā-vṛttas"[13]

Citations

[edit]
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