Fibonacci sequence

In mathematics, the Fibonacci sequence is a sequence in which each element is the sum of the two elements that precede it. Numbers that are part of the Fibonacci sequence are known as Fibonacci numbers, commonly denoted Fn . Many writers begin the sequence with 0 and 1, although some authors start it from 1 and 1 and some (as did Fibonacci) from 1 and 2. Starting from 0 and 1, the sequence begins
The Fibonacci numbers were first described in Indian mathematics as early as 200 BC in work by Pingala on enumerating possible patterns of Sanskrit poetry formed from syllables of two lengths. They are named after the Italian mathematician Leonardo of Pisa, also known as Fibonacci, who introduced the sequence to Western European mathematics in his 1202 book Liber Abaci.
Fibonacci numbers appear unexpectedly often in mathematics, so much so that there is an entire journal dedicated to their study, the Fibonacci Quarterly. Applications of Fibonacci numbers include computer algorithms such as the Fibonacci search technique and the Fibonacci heap data structure, and graphs called Fibonacci cubes used for interconnecting parallel and distributed systems. They also appear in biological settings, such as branching in trees, the arrangement of leaves on a stem, the fruit sprouts of a pineapple, the flowering of an artichoke, and the arrangement of a pine cone's bracts, though they do not occur in all species.
Fibonacci numbers are also strongly related to the golden ratio: Binet's formula expresses the n-th Fibonacci number in terms of n and the golden ratio, and implies that the ratio of two consecutive Fibonacci numbers tends to the golden ratio as n increases. Fibonacci numbers are also closely related to Lucas numbers, which obey the same recurrence relation and with the Fibonacci numbers form a complementary pair of Lucas sequences.
The Fibonacci numbers may be defined by the recurrence relation F 0 = 0 , F 1 = 1 , {\displaystyle F_{0}=0,\quad F_{1}=1,} and F n = F n − 1 + F n − 2 {\displaystyle F_{n}=F_{n-1}+F_{n-2}} for n > 1.
Under some older definitions, the value F 0 = 0 {\displaystyle F_{0}=0} is omitted, so that the sequence starts with F 1 = F 2 = 1 , {\displaystyle F_{1}=F_{2}=1,} and the recurrence F n = F n − 1 + F n − 2 {\displaystyle F_{n}=F_{n-1}+F_{n-2}} is valid for n > 2.
The first 20 Fibonacci numbers Fn are:
The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody. 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 duration m units is Fm+1.
Knowledge of the Fibonacci sequence was expressed as early as Pingala (c. 450 BC–200 BC). Singh cites Pingala's cryptic formula misrau cha ("the two are mixed") and scholars who interpret it in context as saying that the number of patterns for m beats (Fm+1) is obtained by adding one [S] to the Fm cases and one [L] to the Fm−1 cases. Bharata Muni also expresses knowledge of the sequence in the Natya Shastra (c. 100 BC–c. 350 AD). However, the clearest exposition of the sequence arises in the work of Virahanka (c. 700 AD), whose own work is lost, but is available in a quotation by Gopala (c. 1135):
Hemachandra (c. 1150) is credited with knowledge of the sequence as well, writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."
The Fibonacci sequence first appears in the book Liber Abaci (The Book of Calculation, 1202) by Fibonacci, where it is used to calculate the growth of rabbit populations. 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 rabbit math problem: how many pairs will there be in one year?
At the end of the n-th month, the number of pairs of rabbits is equal to the number of mature pairs (that is, the number of pairs in month n – 2) plus the number of pairs alive last month (month n – 1). The number in the n-th month is the n-th Fibonacci number.
The name "Fibonacci sequence" was first used by the 19th-century number theorist Édouard Lucas.
Like every sequence defined by a homogeneous linear recurrence with constant coefficients, the Fibonacci numbers have a closed-form expression. It has become known as Binet's formula, named after French mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre and Daniel Bernoulli:
F n = φ n − ψ n φ − ψ = φ n − ψ n 5 , {\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}={\frac {\varphi ^{n}-\psi ^{n}}{\sqrt {5}}},}
where
φ = 1 + 5 2 ≈ 1.61803 39887 … {\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803\,39887\ldots }
is the golden ratio, and ψ {\displaystyle \psi } is its conjugate:
ψ = 1 − 5 2 = 1 − φ = − 1 φ ≈ − 0.61803 39887 … . {\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}=1-\varphi =-{1 \over \varphi }\approx -0.61803\,39887\ldots .}
Since ψ = − φ − 1 {\displaystyle \psi =-\varphi ^{-1}} , this formula can also be written as
F n = φ n − ( − φ ) − n 5 = φ n − ( − φ ) − n 2 φ − 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, note that φ {\displaystyle \varphi } and ψ {\displaystyle \psi } are both solutions of the equation x 2 = x + 1 {\textstyle x^{2}=x+1} and thus x n = x n − 1 + x n − 2 , {\displaystyle x^{n}=x^{n-1}+x^{n-2},} so the powers of φ {\displaystyle \varphi } and ψ {\displaystyle \psi } satisfy the Fibonacci recursion. In other words,
φ n = φ n − 1 + φ n − 2 , ψ n = ψ n − 1 + ψ n − 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 values a and b, the sequence defined by
U n = a φ n + b ψ n {\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}
satisfies the same recurrence,
U n = a φ n + b ψ n = a ( φ n − 1 + φ n − 2 ) + b ( ψ n − 1 + ψ n − 2 ) = a φ n − 1 + b ψ n − 1 + a φ n − 2 + b ψ n − 2 = U n − 1 + U n − 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}}}
If a and b are chosen so that U0 = 0 and U1 = 1 then the resulting sequence Un must be the Fibonacci sequence. This is the same as requiring a and b satisfy the system of equations:
{ a + b = 0 φ a + ψ b = 1 {\displaystyle \left\{{\begin{aligned}a+b&=0\\\varphi a+\psi b&=1\end{aligned}}\right.}
which has solution
a = 1 φ − ψ = 1 5 , b = − a , {\displaystyle a={\frac {1}{\varphi -\psi }}={\frac {1}{\sqrt {5}}},\quad b=-a,}
producing the required formula.
Taking the starting values U0 and U1 to be arbitrary constants, a more general solution is:
U n = a φ n + b ψ n {\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}
where
a = U 1 − U 0 ψ 5 , b = U 0 φ − U 1 5 . {\displaystyle {\begin{aligned}a&={\frac {U_{1}-U_{0}\psi }{\sqrt {5}}},\\[3mu]b&={\frac {U_{0}\varphi -U_{1}}{\sqrt {5}}}.\end{aligned}}}
Since | ψ n 5 | < 1 2 {\textstyle \left|{\frac {\psi ^{n}}{\sqrt {5}}}\right|<{\frac {1}{2}}} for all n ≥ 0, the number Fn is the closest integer to φ n 5 {\displaystyle {\frac {\varphi ^{n}}{\sqrt {5}}}} . Therefore, it can be found by rounding, using the nearest integer function: F n = ⌊ φ n 5 ⌉ , 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 as n grows, being less than 0.1 for n ≥ 4, and less than 0.01 for n ≥ 8. This formula is easily inverted to find an index of a Fibonacci number F: n ( F ) = ⌊ log φ ⁡ 5 F ⌉ , F ≥ 1. {\displaystyle n(F)=\left\lfloor \log _{\varphi }{\sqrt {5}}F\right\rceil ,\ F\geq 1.}
Instead using the floor function gives the largest index of a Fibonacci number that is not greater than F: n l a r g e s t ( F ) = ⌊ log φ ⁡ 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,} where log φ ⁡ ( x ) = ln ⁡ ( x ) / ln ⁡ ( φ ) = log 10 ⁡ ( x ) / log 10 ⁡ ( 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 } , and log 10 ⁡ ( φ ) = 0.208987 … {\displaystyle \log _{10}(\varphi )=0.208987\ldots } .
Since Fn is asymptotic to φ n / 5 {\displaystyle \varphi ^{n}/{\sqrt {5}}} , the number of digits in Fn is asymptotic to n log 10 ⁡ φ ≈ 0.2090 n {\displaystyle n\log _{10}\varphi \approx 0.2090\,n} . As a consequence, for every integer d > 1 there are either 4 or 5 Fibonacci numbers with d decimal digits.
More generally, in the base b representation, the number of digits in Fn is asymptotic to n log b ⁡ φ = n log ⁡ φ log ⁡ b . {\displaystyle n\log _{b}\varphi ={\frac {n\log \varphi }{\log b}}.}
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 these ratios approach the golden ratio φ : {\displaystyle \varphi \colon } lim n → ∞ F n + 1 F n = φ . {\displaystyle \lim _{n\to \infty }{\frac {F_{n+1}}{F_{n}}}=\varphi .}
This convergence holds regardless of the starting values U 0 {\displaystyle U_{0}} and U 1 {\displaystyle U_{1}} , unless U 1 = − U 0 / φ {\displaystyle U_{1}=-U_{0}/\varphi } . This can be verified using Binet'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, lim n → ∞ F n + m F n = φ m {\displaystyle \lim _{n\to \infty }{\frac {F_{n+m}}{F_{n}}}=\varphi ^{m}} , because the ratios between consecutive Fibonacci numbers approaches φ {\displaystyle \varphi } .
Since the golden ratio satisfies the equation φ 2 = φ + 1 , {\displaystyle \varphi ^{2}=\varphi +1,}
this expression can be used to decompose higher powers φ n {\displaystyle \varphi ^{n}} as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of φ {\displaystyle \varphi } and 1. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients: φ n = F n φ + F n − 1 . {\displaystyle \varphi ^{n}=F_{n}\varphi +F_{n-1}.} This equation can be proved by induction on n ≥ 1: φ n + 1 = ( F n φ + F n − 1 ) φ = F n φ 2 + F n − 1 φ = F n ( φ + 1 ) + F n − 1 φ = ( F n + F n − 1 ) φ + F n = F n + 1 φ + F n . {\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}.} For ψ = − 1 / φ {\displaystyle \psi =-1/\varphi } , it is also the case that ψ 2 = ψ + 1 {\displaystyle \psi ^{2}=\psi +1} and it is also the case that ψ n = F n ψ + F n − 1 . {\displaystyle \psi ^{n}=F_{n}\psi +F_{n-1}.}
These expressions are also true for n < 1 if the Fibonacci sequence Fn is extended to negative integers using the Fibonacci rule F n = F n + 2 − F n + 1 . {\displaystyle F_{n}=F_{n+2}-F_{n+1}.}
Binet's formula provides a proof that a positive integer x is a Fibonacci number if and only if at least one of 5 x 2 + 4 {\displaystyle 5x^{2}+4} or 5 x 2 − 4 {\displaystyle 5x^{2}-4} is a perfect square. This is because Binet's formula, which can be written as F n = ( φ n − ( − 1 ) n φ − n ) / 5 {\displaystyle F_{n}=(\varphi ^{n}-(-1)^{n}\varphi ^{-n})/{\sqrt {5}}} , can be multiplied by 5 φ n {\displaystyle {\sqrt {5}}\varphi ^{n}} and solved as a quadratic equation in φ n {\displaystyle \varphi ^{n}} via the quadratic formula:
φ n = F n 5 ± 5 F n 2 + 4 ( − 1 ) n 2 . {\displaystyle \varphi ^{n}={\frac {F_{n}{\sqrt {5}}\pm {\sqrt {5{F_{n}}^{2}+4(-1)^{n}}}}{2}}.}
Comparing this to φ n = F n φ + F n − 1 = ( F n 5 + F n + 2 F n − 1 ) / 2 {\displaystyle \varphi ^{n}=F_{n}\varphi +F_{n-1}=(F_{n}{\sqrt {5}}+F_{n}+2F_{n-1})/2} , it follows that
In particular, the left-hand side is a perfect square.
A 2-dimensional system of linear difference equations that describes the Fibonacci sequence is
( F k + 2 F k + 1 ) = ( 1 1 1 0 ) ( F k + 1 F k ) {\displaystyle {F_{k+2} \choose F_{k+1}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{F_{k+1} \choose F_{k}}} alternatively denoted F → k + 1 = A F → k , {\displaystyle {\vec {F}}_{k+1}=\mathbf {A} {\vec {F}}_{k},}
which yields F → n = A n F → 0 {\displaystyle {\vec {F}}_{n}=\mathbf {A} ^{n}{\vec {F}}_{0}} . The eigenvalues of the matrix A are φ = 1 2 ( 1 + 5 ) {\displaystyle \varphi ={\tfrac {1}{2}}{\bigl (}1+{\sqrt {5}}~\!{\bigr )}} and ψ = − φ − 1 = 1 2 ( 1 − 5 ) {\displaystyle \psi =-\varphi ^{-1}={\tfrac {1}{2}}{\bigl (}1-{\sqrt {5}}~\!{\bigr )}} corresponding to the respective eigenvectors μ → = ( φ 1 ) , ν → = ( − φ − 1 1 ) . {\displaystyle {\vec {\mu }}={\varphi \choose 1},\quad {\vec {\nu }}={-\varphi ^{-1} \choose 1}.}
As the initial value is F → 0 = ( 1 0 ) = 1 5 μ → − 1 5 ν → , {\displaystyle {\vec {F}}_{0}={1 \choose 0}={\frac {1}{\sqrt {5}}}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}{\vec {\nu }},}
it follows that the nth element is F → n = 1 5 A n μ → − 1 5 A n ν → = 1 5 φ n μ → − 1 5 ( − φ ) − n ν → = 1 5 ( 1 + 5 2 ) n ( φ 1 ) − 1 5 ( 1 − 5 2 ) n ( − φ − 1 1 ) . {\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, the nth element in the Fibonacci series may be read off directly as a closed-form expression: F n = 1 5 ( 1 + 5 2 ) n − 1 5 ( 1 − 5 2 ) 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 by diagonalization of A through use of its eigendecomposition:
A = S Λ S − 1 , A n = S Λ n S − 1 , {\displaystyle {\begin{aligned}A&=S\Lambda S^{-1},\\[3mu]A^{n}&=S\Lambda ^{n}S^{-1},\end{aligned}}}
where
Λ = ( φ 0 0 − φ − 1 ) , S = ( φ − φ − 1 1 1 ) . {\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 the nth element in the Fibonacci series is therefore given by
( F n + 1 F n ) = A n ( F 1 F 0 ) = S Λ n S − 1 ( F 1 F 0 ) = S ( φ n 0 0 ( − φ ) − n ) S − 1 ( F 1 F 0 ) = ( φ − φ − 1 1 1 ) ( φ n 0 0 ( − φ ) − n ) 1 5 ( 1 φ − 1 − 1 φ ) ( 1 0 ) , {\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 yields F n = φ n − ( − φ ) − n 5 . {\displaystyle F_{n}={\cfrac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}.}
The matrix A 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 φ:
φ = 1 + 1 1 + 1 1 + 1 1 + ⋱ . {\displaystyle \varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}}}.}
The convergents of the continued fraction for φ are ratios of successive Fibonacci numbers: φn = Fn+1 / Fn is the n-th convergent, and the (n + 1)-st convergent can be found from the recurrence relation φn+1 = 1 + 1 / φn. 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:
( 1 1 1 0 ) n = ( F n + 1 F n F n F n − 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 given n, this matrix can be computed in O(log n) arithmetic operations, using the exponentiation by squaring method.
Taking the determinant of both sides of this equation yields Cassini's identity, ( − 1 ) n = F n + 1 F n − 1 − F n 2 . {\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-{F_{n}}^{2}.}
Moreover, since AA = A for any square matrix A, the following identities can be derived (they are obtained from two different coefficients of the matrix product, and one may easily deduce the second one from the first one by changing n into n + 1), F m F n + F m − 1 F n − 1 = F m + n − 1 , F m F n + 1 + F m − 1 F n = F m + 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, with m = n, F 2 n − 1 = F n 2 + F n − 1 2 F 2 n − 1 = ( F n − 1 + F n + 1 ) F n = ( 2 F n − 1 + F n ) F n = ( 2 F n + 1 − F n ) F n . {\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 numbers recursively in O(log n) arithmetic operations. This matches the time for computing the n-th Fibonacci number from the closed-form matrix formula, but with fewer redundant steps if one avoids recomputing an already computed Fibonacci number (recursion with memoization).
Most identities involving Fibonacci numbers can be proved using combinatorial arguments using the fact that F n {\displaystyle F_{n}} can be interpreted as the number of (possibly empty) sequences of 1s and 2s whose sum is n − 1 {\displaystyle n-1} . This can be taken as the definition of F n {\displaystyle F_{n}} with the conventions F 0 = 0 {\displaystyle F_{0}=0} , meaning no such sequence exists whose sum is −1, and F 1 = 1 {\displaystyle F_{1}=1} , meaning the empty sequence "adds up" to 0. In the following, | . . . | {\displaystyle |{...}|} is the cardinality of a set:
In this manner the recurrence relation F n = F n − 1 + F n − 2 {\displaystyle F_{n}=F_{n-1}+F_{n-2}} may be understood by dividing the F n {\displaystyle F_{n}} sequences into two non-overlapping sets where all sequences either begin with 1 or 2: F n = | { ( 1 , . . . ) , ( 1 , . . . ) , . . . } | + | { ( 2 , . . . ) , ( 2 , . . . ) , . . . } | {\displaystyle F_{n}=|\{(1,...),(1,...),...\}|+|\{(2,...),(2,...),...\}|} Excluding the first element, the remaining terms in each sequence sum to n − 2 {\displaystyle n-2} or n − 3 {\displaystyle n-3} and the cardinality of each set is F n − 1 {\displaystyle F_{n-1}} or F n − 2 {\displaystyle F_{n-2}} giving a total of F n − 1 + F n − 2 {\displaystyle F_{n-1}+F_{n-2}} sequences, showing this is equal to F n {\displaystyle F_{n}} .
In a similar manner it may be shown that the sum of the first Fibonacci numbers up to the n-th is equal to the (n + 2)-th Fibonacci number minus 1. In symbols: ∑ i = 1 n F i = F n + 2 − 1 {\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1}
This may be seen by dividing all sequences summing to n + 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
... where the last two terms have the value F 1 = 1 {\displaystyle F_{1}=1} . From this it follows that ∑ i = 1 n F i = F n + 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 = 0 n − 1 F 2 i + 1 = F 2 n {\displaystyle \sum _{i=0}^{n-1}F_{2i+1}=F_{2n}} and ∑ i = 1 n F 2 i = F 2 n + 1 − 1. {\displaystyle \sum _{i=1}^{n}F_{2i}=F_{2n+1}-1.} In words, the sum of the first Fibonacci numbers with odd index up to F 2 n − 1 {\displaystyle F_{2n-1}} is the (2n)-th Fibonacci number, and the sum of the first Fibonacci numbers with even index up to F 2 n {\displaystyle F_{2n}} is the (2n + 1)-th Fibonacci number minus 1.
A different trick may be used to prove ∑ i = 1 n F i 2 = F n F n + 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 to F n {\displaystyle F_{n}} is the product of the n-th and (n + 1)-th Fibonacci numbers. To see this, begin with a Fibonacci rectangle of size F n × F n + 1 {\displaystyle F_{n}\times F_{n+1}} and decompose it into squares of size F n , F n − 1 , . . . , F 1 {\displaystyle F_{n},F_{n-1},...,F_{1}} ; from this the identity follows by comparing areas:
The sequence ( F n ) n ∈ N {\displaystyle (F_{n})_{n\in \mathbb {N} }} is also considered using the symbolic method. More precisely, this sequence corresponds to a specifiable combinatorial class. The specification of this sequence is Seq ⁡ ( Z + Z 2 ) {\displaystyle \operatorname {Seq} ({\mathcal {Z+Z^{2}}})} . Indeed, as stated above, the n {\displaystyle n} -th Fibonacci number equals the number of combinatorial compositions (ordered partitions) of n − 1 {\displaystyle n-1} using terms 1 and 2.
It follows that the ordinary generating function of the Fibonacci sequence, ∑ i = 0 ∞ F i z i {\displaystyle \sum _{i=0}^{\infty }F_{i}z^{i}} , is the rational function z 1 − z − z 2 . {\displaystyle {\frac {z}{1-z-z^{2}}}.}
Fibonacci identities often can be easily proved using mathematical induction.
For example, reconsider ∑ i = 1 n F i = F n + 2 − 1. {\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1.} Adding F n + 1 {\displaystyle F_{n+1}} to both sides gives
and so we have the formula for n + 1 {\displaystyle n+1} ∑ i = 1 n + 1 F i = F n + 3 − 1 {\displaystyle \sum _{i=1}^{n+1}F_{i}=F_{n+3}-1}
Similarly, add F n + 1 2 {\displaystyle {F_{n+1}}^{2}} to both sides of ∑ i = 1 n F i 2 = F n F n + 1 {\displaystyle \sum _{i=1}^{n}F_{i}^{2}=F_{n}F_{n+1}} to give ∑ i = 1 n F i 2 + F n + 1 2 = F n + 1 ( F n + F n + 1 ) {\displaystyle \sum _{i=1}^{n}F_{i}^{2}+{F_{n+1}}^{2}=F_{n+1}\left(F_{n}+F_{n+1}\right)} ∑ i = 1 n + 1 F i 2 = F n + 1 F n + 2 {\displaystyle \sum _{i=1}^{n+1}F_{i}^{2}=F_{n+1}F_{n+2}}
The Binet formula is 5 F n = φ n − ψ n . {\displaystyle {\sqrt {5}}F_{n}=\varphi ^{n}-\psi ^{n}.} This can be used to prove Fibonacci identities.
For example, to prove that ∑ i = 1 n F i = F n + 2 − 1 {\textstyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1} note that the left hand side multiplied by 5 {\displaystyle {\sqrt {5}}} becomes 1 + φ + φ 2 + ⋯ + φ n − ( 1 + ψ + ψ 2 + ⋯ + ψ n ) = φ n + 1 − 1 φ − 1 − ψ n + 1 − 1 ψ − 1 = φ n + 1 − 1 − ψ − ψ n + 1 − 1 − φ = − φ n + 2 + φ + ψ n + 2 − ψ φ ψ = φ n + 2 − ψ n + 2 − ( φ − ψ ) = 5 ( F n + 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 facts φ ψ = − 1 {\textstyle \varphi \psi =-1} and φ − ψ = 5 {\textstyle \varphi -\psi ={\sqrt {5}}} to simplify the equations.
Numerous other identities can be derived using various methods. Here are some of them:
Cassini's identity states that F n 2 − F n + 1 F n − 1 = ( − 1 ) n − 1 {\displaystyle {F_{n}}^{2}-F_{n+1}F_{n-1}=(-1)^{n-1}} Catalan's identity is a generalization: F n 2 − F n + r F n − r = ( − 1 ) n − r F r 2 {\displaystyle {F_{n}}^{2}-F_{n+r}F_{n-r}=(-1)^{n-r}{F_{r}}^{2}}
F m F n + 1 − F m + 1 F n = ( − 1 ) n F m − n {\displaystyle F_{m}F_{n+1}-F_{m+1}F_{n}=(-1)^{n}F_{m-n}} F 2 n = F n + 1 2 − F n − 1 2 = F n ( F n + 1 + F n − 1 ) = F n L n {\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}} where Ln is the n-th Lucas number. The last is an identity for doubling n; other identities of this type are F 3 n = 2 F n 3 + 3 F n F n + 1 F n − 1 = 5 F n 3 + 3 ( − 1 ) n F n {\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.
F 3 n + 1 = F n + 1 3 + 3 F n + 1 F n 2 − F n 3 {\displaystyle F_{3n+1}={F_{n+1}}^{3}+3F_{n+1}{F_{n}}^{2}-{F_{n}}^{3}} F 3 n + 2 = F n + 1 3 + 3 F n + 1 2 F n + F n 3 {\displaystyle F_{3n+2}={F_{n+1}}^{3}+3{F_{n+1}}^{2}F_{n}+{F_{n}}^{3}} F 4 n = 4 F n F n + 1 ( F n + 1 2 + 2 F n 2 ) − 3 F n 2 ( F n 2 + 2 F n + 1 2 ) {\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 using lattice reduction, and are useful in setting up the special number field sieve to factorize a Fibonacci number.
More generally,
F k n + c = ∑ i = 0 k ( k i ) F c − i F n i F n + 1 k − 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
F k n + c = ∑ i = 0 k ( k i ) F c + i F n i F n − 1 k − i . {\displaystyle F_{kn+c}=\sum _{i=0}^{k}{k \choose i}F_{c+i}{F_{n}}^{i}{F_{n-1}}^{k-i}.}
Putting k = 2 in this formula, one gets again the formulas of the end of above section Matrix form.
The generating function of the Fibonacci sequence is the power series
s ( z ) = ∑ k = 0 ∞ F k z k = 0 + z + z 2 + 2 z 3 + 3 z 4 + 5 z 5 + … . {\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 any complex number z {\displaystyle z} satisfying | z | < 1 / φ ≈ 0.618 , {\displaystyle |z|<1/\varphi \approx 0.618,} and its sum has a simple closed form:
s ( z ) = z 1 − z − z 2 . {\displaystyle s(z)={\frac {z}{1-z-z^{2}}}.}
This can be proved by multiplying by ( 1 − z − z 2 ) {\textstyle (1-z-z^{2})} : ( 1 − z − z 2 ) s ( z ) = ∑ k = 0 ∞ F k z k − ∑ k = 0 ∞ F k z k + 1 − ∑ k = 0 ∞ F k z k + 2 = ∑ k = 0 ∞ F k z k − ∑ k = 1 ∞ F k − 1 z k − ∑ k = 2 ∞ F k − 2 z k = 0 z 0 + 1 z 1 − 0 z 1 + ∑ k = 2 ∞ ( F k − F k − 1 − F k − 2 ) z k = 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 involving z k {\displaystyle z^{k}} for k ≥ 2 {\displaystyle k\geq 2} cancel out because of the defining Fibonacci recurrence relation.
The partial fraction decomposition is given by s ( z ) = 1 5 ( 1 1 − φ z − 1 1 − ψ z ) {\displaystyle s(z)={\frac {1}{\sqrt {5}}}\left({\frac {1}{1-\varphi z}}-{\frac {1}{1-\psi z}}\right)} where φ = 1 2 ( 1 + 5 ) {\textstyle \varphi ={\tfrac {1}{2}}\left(1+{\sqrt {5}}\right)} is the golden ratio and ψ = 1 2 ( 1 − 5 ) {\displaystyle \psi ={\tfrac {1}{2}}\left(1-{\sqrt {5}}\right)} is its conjugate.
The related function z ↦ − s ( − 1 / z ) {\textstyle z\mapsto -s\left(-1/z\right)} is the generating function for the negafibonacci numbers, and s ( z ) {\displaystyle s(z)} satisfies the functional equation
s ( z ) = s ( − 1 z ) . {\displaystyle s(z)=s\!\left(-{\frac {1}{z}}\right).}
Using z {\displaystyle z} equal to any of 0.01, 0.001, 0.0001, etc. lays out the first Fibonacci numbers in the decimal expansion of s ( z ) {\displaystyle s(z)} . For example, s ( 0.001 ) = 0.001 0.998999 = 1000 998999 = 0.001001002003005008013021 … . {\displaystyle s(0.001)={\frac {0.001}{0.998999}}={\frac {1000}{998999}}=0.001001002003005008013021\ldots .}
Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of theta functions. For example, the sum of every odd-indexed reciprocal Fibonacci number can be written as ∑ k = 1 ∞ 1 F 2 k − 1 = 5 4 ϑ 2 ( 0 , 3 − 5 2 ) 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 ∞ 1 F k 2 = 5 24 ( ϑ 2 ( 0 , 3 − 5 2 ) 4 − ϑ 4 ( 0 , 3 − 5 2 ) 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 ∞ 1 1 + F 2 k − 1 = 5 2 , {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{1+F_{2k-1}}}={\frac {\sqrt {5}}{2}},}
and there is a nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio, ∑ k = 1 ∞ ( − 1 ) k + 1 ∑ j = 1 k F j 2 = 5 − 1 2 . {\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 ∑ k = 1 ∞ 1 F 2 k = 5 ( L ( ψ 2 ) − L ( ψ 4 ) ) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{2k}}}={\sqrt {5}}\left(L(\psi ^{2})-L(\psi ^{4})\right)} with the Lambert series L ( q ) := ∑ k = 1 ∞ q k 1 − q k , {\displaystyle \textstyle L(q):=\sum _{k=1}^{\infty }{\frac {q^{k}}{1-q^{k}}},} since 1 F 2 k = 5 ( ψ 2 k 1 − ψ 2 k − ψ 4 k 1 − ψ 4 k ) . {\displaystyle \textstyle {\frac {1}{F_{2k}}}={\sqrt {5}}\left({\frac {\psi ^{2k}}{1-\psi ^{2k}}}-{\frac {\psi ^{4k}}{1-\psi ^{4k}}}\right)\!.}
So the reciprocal Fibonacci constant is ∑ k = 1 ∞ 1 F k = ∑ k = 1 ∞ 1 F 2 k − 1 + ∑ k = 1 ∞ 1 F 2 k = 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 proved irrational by Richard André-Jeannin.
Millin's series gives the identity ∑ k = 0 ∞ 1 F 2 k = 7 − 5 2 , {\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 as N tends to infinity: ∑ k = 0 N 1 F 2 k = 3 − F 2 N − 1 F 2 N . {\displaystyle \sum _{k=0}^{N}{\frac {1}{F_{2^{k}}}}=3-{\frac {F_{2^{N}-1}}{F_{2^{N}}}}.}
Every third number of the sequence is even (a multiple of F 3 = 2 {\displaystyle F_{3}=2} ) and, more generally, every k-th number of the sequence is a multiple of Fk. Thus the Fibonacci sequence is an example of a divisibility sequence. In fact, the Fibonacci sequence satisfies the stronger divisibility property gcd ( F a , F b , F c , … ) = F gcd ( a , b , c , … ) {\displaystyle \gcd(F_{a},F_{b},F_{c},\ldots )=F_{\gcd(a,b,c,\ldots )}\,} where gcd is the greatest common divisor function. (This relation is different if a different indexing convention is used, such as the one that starts the sequence with ⁠ F 0 = 1 {\displaystyle F_{0}=1} ⁠ and ⁠ F 1 = 1 {\displaystyle F_{1}=1} ⁠.)
In particular, any three consecutive Fibonacci numbers are pairwise coprime because both F 1 = 1 {\displaystyle F_{1}=1} and F 2 = 1 {\displaystyle F_{2}=1} . That is,
for every n.
Every prime number p divides a Fibonacci number that can be determined by the value of p modulo 5. If p is congruent to 1 or 4 modulo 5, then p divides Fp−1, and if p is congruent to 2 or 3 modulo 5, then, p divides Fp+1. The remaining case is that p = 5, and in this case p divides Fp.
{ p = 5 ⇒ p ∣ F p , p ≡ ± 1 ( mod 5 ) ⇒ p ∣ F p − 1 , p ≡ ± 2 ( mod 5 ) ⇒ p ∣ F p + 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 the Legendre symbol: p ∣ F p − ( 5 p ) . {\displaystyle p\mid F_{p\;-\,\left({\frac {5}{p}}\right)}.}
The above formula can be used as a primality test in the sense that if n ∣ F n − ( 5 n ) , {\displaystyle n\mid F_{n\;-\,\left({\frac {5}{n}}\right)},} where the Legendre symbol has been replaced by the Jacobi symbol, then this is evidence that n is a prime, and if it fails to hold, then n is definitely not a prime. If n is composite and satisfies the formula, then n is a Fibonacci pseudoprime. When m is large – say a 500-bit number – then we can calculate Fm (mod n) efficiently using the matrix form. Thus
( F m + 1 F m F m F m − 1 ) ≡ ( 1 1 1 0 ) m ( mod n ) . {\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 power A is calculated using modular exponentiation, which can be adapted to matrices.
A Fibonacci prime is a Fibonacci number that is prime. The first few are:
Fibonacci primes with thousands of digits have been found, but it is not known whether there are infinitely many.
Fkn is divisible by Fn, so, apart from F4 = 3, any Fibonacci prime must have a prime index. As there are arbitrarily long runs of composite numbers, there are therefore also arbitrarily long runs of composite Fibonacci numbers.
No Fibonacci number greater than F6 = 8 is one greater or one less than a prime number.
The only nontrivial square Fibonacci number is 144. Attila Pethő proved in 2001 that there is only a finite number of perfect power Fibonacci numbers. In 2006, Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only such non-trivial perfect powers.
1, 3, 21, and 55 are the only triangular Fibonacci numbers, which was conjectured by Vern Hoggatt and proved by Luo Ming.
No Fibonacci number can be a perfect number. More generally, no Fibonacci number other than 1 can be multiply perfect, and no ratio of two Fibonacci numbers can be perfect.
With the exceptions of 1, 8 and 144 (F1 = F2, F6 and F12) every Fibonacci number has a prime factor that is not a factor of any smaller Fibonacci number (Carmichael's theorem). As a result, 8 and 144 (F6 and F12) are the only Fibonacci numbers that are the product of other Fibonacci numbers.
The divisibility of Fibonacci numbers by a prime p is related to the Legendre symbol ( p 5 ) {\displaystyle {\bigl (}{\tfrac {p}{5}}{\bigr )}} which is evaluated as follows: ( p 5 ) = { 0 if p = 5 1 if p ≡ ± 1 ( mod 5 ) − 1 if p ≡ ± 2 ( mod 5 ) . {\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}}}
If p is a prime number then F p ≡ ( p 5 ) ( mod p ) and F p − ( p 5 ) ≡ 0 ( mod p ) . {\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}}.}
For example, ( 2 5 ) = − 1 , F 3 = 2 , F 2 = 1 , ( 3 5 ) = − 1 , F 4 = 3 , F 3 = 2 , ( 5 5 ) = 0 , F 5 = 5 , ( 7 5 ) = − 1 , F 8 = 21 , F 7 = 13 , ( 11 5 ) = + 1 , F 10 = 55 , F 11 = 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 prime p such that
F p − ( p 5 ) ≡ 0 ( mod p 2 ) . {\displaystyle F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p^{2}}}.}
Such primes (if there are any) would be called Wall–Sun–Sun primes.
Also, if p ≠ 5 is an odd prime number then: 5 F p ± 1 2 2 ≡ { 1 2 ( 5 ( p 5 ) ± 5 ) ( mod p ) if p ≡ 1 ( mod 4 ) 1 2 ( 5 ( p 5 ) ∓ 3 ) ( mod p ) if p ≡ 3 ( mod 4 ) . {\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 case p ≡ 3 (mod 4) and we have: ( 7 5 ) = − 1 : 1 2 ( 5 ( 7 5 ) + 3 ) = − 1 , 1 2 ( 5 ( 7 5 ) − 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.} F 3 = 2 and F 4 = 3. {\displaystyle F_{3}=2{\text{ and }}F_{4}=3.} 5 F 3 2 = 20 ≡ − 1 ( mod 7 ) and 5 F 4 2 = 45 ≡ − 4 ( mod 7 ) {\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 case p ≡ 3 (mod 4) and we have: ( 11 5 ) = + 1 : 1 2 ( 5 ( 11 5 ) + 3 ) = 4 , 1 2 ( 5 ( 11 5 ) − 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.} F 5 = 5 and F 6 = 8. {\displaystyle F_{5}=5{\text{ and }}F_{6}=8.} 5 F 5 2 = 125 ≡ 4 ( mod 11 ) and 5 F 6 2 = 320 ≡ 1 ( mod 11 ) {\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 case p ≡ 1 (mod 4) and we have: ( 13 5 ) = − 1 : 1 2 ( 5 ( 13 5 ) − 5 ) = − 5 , 1 2 ( 5 ( 13 5 ) + 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.} F 6 = 8 and F 7 = 13. {\displaystyle F_{6}=8{\text{ and }}F_{7}=13.} 5 F 6 2 = 320 ≡ − 5 ( mod 13 ) and 5 F 7 2 = 845 ≡ 0 ( mod 13 ) {\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 case p ≡ 1 (mod 4) and we have: ( 29 5 ) = + 1 : 1 2 ( 5 ( 29 5 ) − 5 ) = 0 , 1 2 ( 5 ( 29 5 ) + 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.} F 14 = 377 and F 15 = 610. {\displaystyle F_{14}=377{\text{ and }}F_{15}=610.} 5 F 14 2 = 710645 ≡ 0 ( mod 29 ) and 5 F 15 2 = 1860500 ≡ 5 ( mod 29 ) {\displaystyle 5{F_{14}}^{2}=710645\equiv 0{\pmod {29}}\;\;{\text{ and }}\;\;5{F_{15}}^{2}=1860500\equiv 5{\pmod {29}}}
For odd n, all odd prime divisors of Fn are congruent to 1 modulo 4, implying that all odd divisors of Fn (as the products of odd prime divisors) are congruent to 1 modulo 4.
For example, F 1 = 1 , F 3 = 2 , F 5 = 5 , F 7 = 13 , F 9 = 34 = 2 ⋅ 17 , F 11 = 89 , F 13 = 233 , F 15 = 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 numbers F(i) for all i < 50000 are collected at the relevant repositories.
If the members of the Fibonacci sequence are taken mod n, the resulting sequence is periodic with period at most 6n. The lengths of the periods for various n form the so-called Pisano periods. Determining a general formula for the Pisano periods is an open problem, which includes as a subproblem a special instance of the problem of finding the multiplicative order of a modular integer or of an element in a finite field. However, for any particular n, the Pisano period may be found as an instance of cycle detection.
The Fibonacci sequence is one of the simplest and earliest known sequences defined by a recurrence relation, and specifically by a linear difference 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 a homogeneous linear difference equation with constant coefficients.
Some specific examples that are close, in some sense, to the Fibonacci sequence include:
The Fibonacci numbers occur as the sums of binomial coefficients in the "shallow" diagonals of Pascal's triangle: F n = ∑ k = 0 ⌊ n − 1 2 ⌋ ( n − k − 1 k ) . {\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 function x 1 − x − x 2 = x + x 2 ( 1 + x ) + x 3 ( 1 + x ) 2 + ⋯ + x k + 1 ( 1 + x ) k + ⋯ = ∑ n = 0 ∞ F n x n {\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 of x n {\displaystyle x^{n}} .
To see how the formula is used, we can arrange the sums by the number of terms present:
which is ( 5 0 ) + ( 4 1 ) + ( 3 2 ) {\displaystyle \textstyle {\binom {5}{0}}+{\binom {4}{1}}+{\binom {3}{2}}} , where we are choosing the positions of k twos from n−k−1 terms.
These numbers also give the solution to certain enumerative problems, the most common of which is that of counting the number of ways of writing a given number n as an ordered sum of 1s and 2s (called compositions); there are Fn+1 ways to do this (equivalently, it's also the number of domino tilings of the 2 × n {\displaystyle 2\times n} rectangle). For example, there are F5+1 = F6 = 8 ways one can climb a staircase of 5 steps, taking one or two steps at a time:
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 applied recursively 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 of binary strings, or equivalently, among the subsets of a given set.
Fibonacci sequences appear in biological settings, such as branching in trees, arrangement of leaves on a stem, the fruitlets of a pineapple, the flowering of artichoke, the arrangement of a pine cone, and the family tree of honeybees. Kepler pointed out the presence of the Fibonacci sequence in nature, using it to explain the (golden ratio-related) pentagonal form of some flowers. Field daisies most often have petals in counts of Fibonacci numbers. In 1830, Karl Friedrich Schimper and Alexander Braun discovered that the parastichies (spiral phyllotaxis) of plants were frequently expressed as fractions involving Fibonacci numbers.
Przemysław Prusinkiewicz advanced the idea that real instances can in part be understood as the expression of certain algebraic constraints on free groups, specifically as certain Lindenmayer grammars.
A model for the pattern of florets in the head of a sunflower was proposed by Helmut Vogel [de] in 1979. This has the form
θ = 2 π φ 2 n , r = c n {\displaystyle \theta ={\frac {2\pi }{\varphi ^{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. Sunflowers and similar flowers most commonly have spirals of florets in clockwise and counter-clockwise directions in the amount of adjacent Fibonacci numbers, typically counted by the outermost range of radii.
Fibonacci numbers also appear in the ancestral pedigrees of bees (which are haplodiploids), according to the following rules:
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 is Fn−1, plus the number of male ancestors, which is Fn−2. This is under the unrealistic assumption that the ancestors at each level are otherwise unrelated.
It has similarly been noticed that the number of possible ancestors on the human X chromosome inheritance line at a given ancestral generation also follows the Fibonacci sequence. A male individual has an X chromosome, which he received from his mother, and a Y chromosome, which he received from his father. The male counts as the "origin" of his own X chromosome ( F 1 = 1 {\displaystyle F_{1}=1} ), and at his parents' generation, his X chromosome came from a single parent ( F 2 = 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 ( F 3 = 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 ( F 4 = 3 {\displaystyle F_{4}=3} ). Five great-great-grandparents contributed to the male descendant's X chromosome ( F 5 = 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 a population founder appears on all lines of the genealogy.)