In mathematics, the Fibonacci sequence is a sequence in which each number is the sum of the two preceding ones. Numbers that are part of the Fibonacci sequence are known as Fibonacci numbers, commonly denoted Шаблон:Nowrap. The sequence commonly starts from 0 and 1, although some authors start the sequence from 1 and 1 or sometimes (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.[2][3][4] 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 Шаблон:Lang.Шаблон:Sfn
Fibonacci numbers are also strongly related to the golden ratio: Binet's formula expresses the Шаблон:Mvar-th Fibonacci number in terms of Шаблон:Mvar and the golden ratio, and implies that the ratio of two consecutive Fibonacci numbers tends to the golden ratio as Шаблон:Mvar 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.
Файл:Fibonacci Spiral.svgThe Fibonacci spiral: an approximation of the golden spiral created by drawing circular arcs connecting the opposite corners of squares in the Fibonacci tiling (see preceding image)
The Fibonacci numbers may be defined by the recurrence relationШаблон:Sfn
<math display=block>F_0=0,\quad F_1= 1,</math>
and
<math display=block>F_n=F_{n-1} + F_{n-2}</math>
for Шаблон:Math.
Under some older definitions, the value <math>F_0 = 0</math> is omitted, so that the sequence starts with <math>F_1=F_2=1,</math> and the recurrence <math>F_n=F_{n-1} + F_{n-2}</math> is valid for Шаблон:Math.Шаблон:SfnШаблон:Sfn
The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody.[3][5]Шаблон:Sfn 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 Шаблон:Mvar units is Шаблон:Math.[4]
Knowledge of the Fibonacci sequence was expressed as early as Pingala (Шаблон:Circa 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 Шаблон:Mvar beats (Шаблон:Math) is obtained by adding one [S] to the Шаблон:Math cases and one [L] to the Шаблон:Math cases.[6]Bharata Muni also expresses knowledge of the sequence in the Natya Shastra (c. 100 BC–c. 350 AD).[7][2]
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):Шаблон:Sfn
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 all mātrā-vṛttas [prosodic combinations].Шаблон:Efn
Hemachandra (c. 1150) is credited with knowledge of the sequence as well,[2] writing that "the sum of the last and the one before the last is the number ... of the next mātrā-vṛtta."Шаблон:Sfn[8]
Europe
Файл:Liber abbaci magliab f124r.jpgA page of Fibonacci's Шаблон:Lang from the Biblioteca 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 book Шаблон:Lang (The Book of Calculation, 1202) by FibonacciШаблон:Sfn[9] where it is used to calculate the growth of rabbit populations.[10][11] 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 puzzle: 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 the Шаблон:Mvar-th month, the number of pairs of rabbits is equal to the number of mature pairs (that is, the number of pairs in month Шаблон:Math) plus the number of pairs alive last month (month Шаблон:Math). The number in the Шаблон:Mvar-th month is the Шаблон:Mvar-th Fibonacci number.[12]
The name "Fibonacci sequence" was first used by the 19th-century number theorist Édouard Lucas.[13]
Файл:Fibonacci Rabbits.svgIn a growing idealized population, the number of rabbit pairs form the Fibonacci sequence. At the end of the nth month, the number of pairs is equal to Fn.
To see the relation between the sequence and these constants,Шаблон:Sfn note that Шаблон:Mvar and Шаблон:Mvar are both solutions of the equation <math display=inline>x^2 = x + 1</math> and thus <math>x^n = x^{n-1} + x^{n-2},</math> so the powers of Шаблон:Mvar and Шаблон:Mvar satisfy the Fibonacci recursion. In other words,
Since
<math display=inline>\left|\frac{\psi^{n}}{\sqrt 5}\right| < \frac{1}{2}</math> for all Шаблон:Math, the number Шаблон:Math is the closest integer to <math>\frac{\varphi^n}{\sqrt 5}</math>. Therefore, it can be found by rounding, using the nearest integer function:
<math display=block>F_n=\left\lfloor\frac{\varphi^n}{\sqrt 5}\right\rceil,\ n \geq 0.</math>
In fact, the rounding error is very small, being less than 0.1 for Шаблон:Math, and less than 0.01 for Шаблон:Math. This formula is easily inverted to find an index of a Fibonacci number Шаблон:Mvar:
<math display=block>n(F) = \left\lfloor \log_\varphi \sqrt{5}F\right\rceil,\ F \geq 1.</math>
Instead using the floor function gives the largest index of a Fibonacci number that is not greater than Шаблон:Mvar:
<math display=block>n_{\mathrm{largest}}(F) = \left\lfloor \log_\varphi \sqrt{5}(F+1/2)\right\rfloor,\ F \geq 0,</math>
where <math>\log_\varphi(x) = \ln(x)/\ln(\varphi) = \log_{10}(x)/\log_{10}(\varphi)</math>, <math>\ln(\varphi) = 0.481211\ldots</math>,[15] and <math>\log_{10}(\varphi) = 0.208987\ldots</math>.[16]
Magnitude
Since Fn is asymptotic to <math>\varphi^n/\sqrt5</math>, the number of digits in Шаблон:Math is asymptotic to <math>n\log_{10}\varphi\approx 0.2090\, n</math>. As a consequence, for every integer Шаблон:Math there are either 4 or 5 Fibonacci numbers with Шаблон:Mvar decimal digits.
More generally, in the baseШаблон:Mvar representation, the number of digits in Шаблон:Math is asymptotic to <math>n\log_b\varphi = \frac{n \log \varphi}{\log b}.</math>
Limit of consecutive quotients
Johannes Kepler observed that the ratio of consecutive Fibonacci numbers converges. He wrote that "as 5 is to 8 so is 8 to 13, practically, and as 8 is to 13, so is 13 to 21 almost", and concluded that these ratios approach the golden ratio <math>\varphi\colon </math> [17][18]
<math display=block>\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi.</math>
This convergence holds regardless of the starting values <math>U_0</math> and <math>U_1</math>, unless <math>U_1 = -U_0/\varphi</math>. 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 terms in this sequence shows the same convergence towards the golden ratio.
In general, <math>\lim_{n\to\infty}\frac{F_{n+m}}{F_n}=\varphi^m
</math>, because the ratios between consecutive Fibonacci numbers approaches <math>\varphi</math>.
Since the golden ratio satisfies the equation
<math display=block>\varphi^2 = \varphi + 1,</math>
this expression can be used to decompose higher powers <math>\varphi^n</math> as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of <math>\varphi</math> and 1. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients:
<math display=block>\varphi^n = F_n\varphi + F_{n-1}.</math>
This equation can be proved by induction on Шаблон:Math:
<math display=block>\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.</math>
For <math>\psi = -1/\varphi</math>, it is also the case that <math>\psi^2 = \psi + 1</math> and it is also the case that
<math display=block>\psi^n = F_n\psi + F_{n-1}.</math>
These expressions are also true for Шаблон:Math if the Fibonacci sequence Fn is extended to negative integers using the Fibonacci rule <math>F_n = F_{n+2} - F_{n+1}.</math>
Identification
Binet's formula provides a proof that a positive integer Шаблон:Mvar is a Fibonacci number if and only if at least one of <math>5x^2+4</math> or <math>5x^2-4</math> is a perfect square.[19] This is because Binet's formula, which can be written as <math>F_n = (\varphi^n - (-1)^n \varphi^{-n}) / \sqrt{5}</math>, can be multiplied by <math>\sqrt{5} \varphi^n</math> and solved as a quadratic equation in <math>\varphi^n</math> via the quadratic formula:
which yields <math>\vec F_n = \mathbf{A}^n \vec F_0</math>. The eigenvalues of the matrixШаблон:Math are <math>\varphi=\frac12(1+\sqrt5)</math> and <math>\psi=-\varphi^{-1}=\frac12(1-\sqrt5)</math> corresponding to the respective eigenvectors
<math display=block>\vec \mu={\varphi \choose 1}</math>
and
<math display=block>\vec\nu={-\varphi^{-1} \choose 1}.</math>
As the initial value is
<math display=block>\vec F_0={1 \choose 0}=\frac{1}{\sqrt{5}}\vec{\mu}-\frac{1}{\sqrt{5}}\vec{\nu},</math>
it follows that the Шаблон:Mvarth term is
<math display=block>\begin{align}\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{align}</math>
From this, the Шаблон:Mvarth element in the Fibonacci series
may be read off directly as a closed-form expression:
<math display=block>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.</math>
Equivalently, the same computation may be performed by diagonalization of Шаблон:Math through use of its eigendecomposition:
<math display=block>\begin{align} A & = S\Lambda S^{-1} ,\\
A^n & = S\Lambda^n S^{-1},
\end{align}</math>
where <math>\Lambda=\begin{pmatrix} \varphi & 0 \\ 0 & -\varphi^{-1} \end{pmatrix}</math> and <math>S=\begin{pmatrix} \varphi & -\varphi^{-1} \\ 1 & 1 \end{pmatrix}.</math>
The closed-form expression for the Шаблон:Mvarth element in the Fibonacci series is therefore given by
The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for Шаблон:Mvar, and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1. The matrix representation gives the following closed-form expression for the Fibonacci numbers:
These last two identities provide a way to compute Fibonacci numbers recursively in Шаблон:Math arithmetic operations. This matches the time for computing the Шаблон:Mvar-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).[20]
Combinatorial identities
Combinatorial proofs
Most identities involving Fibonacci numbers can be proved using combinatorial arguments using the fact that <math>F_n</math> can be interpreted as the number of (possibly empty) sequences of 1s and 2s whose sum is <math>n-1</math>. This can be taken as the definition of <math>F_n</math> with the conventions <math>F_0 = 0</math>, meaning no such sequence exists whose sum is −1, and <math>F_1 = 1</math>, meaning the empty sequence "adds up" to 0. In the following, <math>|{...}|</math> is the cardinality of a set:
In this manner the recurrence relation
<math display=block>F_n = F_{n-1} + F_{n-2}</math>
may be understood by dividing the <math>F_n</math> sequences into two non-overlapping sets where all sequences either begin with 1 or 2:
<math display=block>F_n = |\{(1,...),(1,...),...\}| + |\{(2,...),(2,...),...\}|</math>
Excluding the first element, the remaining terms in each sequence sum to <math>n-2</math> or <math>n-3</math> and the cardinality of each set is <math>F_{n-1}</math> or <math>F_{n-2}</math> giving a total of <math>F_{n-1}+F_{n-2}</math> sequences, showing this is equal to <math>F_n</math>.
In a similar manner it may be shown that the sum of the first Fibonacci numbers up to the Шаблон:Mvar-th is equal to the Шаблон:Math-th Fibonacci number minus 1.Шаблон:Sfn In symbols:
<math display=block>\sum_{i=1}^n F_i = F_{n+2} - 1</math>
This may be seen by dividing all sequences summing to <math>n+1</math> based on the location of the first 2. Specifically, each set consists of those sequences that start <math>(2,...), (1,2,...), ..., </math> until the last two sets <math>\{(1,1,...,1,2)\}, \{(1,1,...,1)\}</math> 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 <math>F_1 = 1</math>. From this it follows that <math>\sum_{i=1}^n F_i = F_{n+2}-1</math>.
A similar argument, grouping the sums by the position of the first 1 rather than the first 2 gives two more identities:
<math display=block>\sum_{i=0}^{n-1} F_{2 i+1} = F_{2 n}</math>
and
<math display=block>\sum_{i=1}^{n} F_{2 i} = F_{2 n+1}-1.</math>
In words, the sum of the first Fibonacci numbers with odd index up to <math>F_{2 n-1}</math> is the Шаблон:Math-th Fibonacci number, and the sum of the first Fibonacci numbers with even index up to <math>F_{2 n}</math> is the Шаблон:Math-th Fibonacci number minus 1.[21]
A different trick may be used to prove
<math display=block>\sum_{i=1}^n F_i^2 = F_n F_{n+1}</math>
or in words, the sum of the squares of the first Fibonacci numbers up to <math>F_n</math> is the product of the Шаблон:Mvar-th and Шаблон:Math-th Fibonacci numbers. To see this, begin with a Fibonacci rectangle of size <math>F_n \times F_{n+1}</math> and decompose it into squares of size <math>F_n, F_{n-1}, ..., F_1</math>; from this the identity follows by comparing areas:
The sequence <math>(F_n)_{n\in\mathbb N}</math> is also considered using the symbolic method.[22] More precisely, this sequence corresponds to a specifiable combinatorial class. The specification of this sequence is <math>\operatorname{Seq}(\mathcal{Z+Z^2})</math>. Indeed, as stated above, the <math>n</math>-th Fibonacci number equals the number of combinatorial compositions (ordered partitions) of <math>n-1</math> using terms 1 and 2.
and so we have the formula for <math>n+1</math>
<math display=block>\sum_{i=1}^{n+1} F_i = F_{n+3} - 1</math>
Similarly, add <math>{F_{n+1}}^2</math> to both sides of
<math display=block>\sum_{i=1}^n F_i^2 = F_n F_{n+1}</math>
to give
<math display=block>\sum_{i=1}^n F_i^2 + {F_{n+1}}^2 = F_{n+1}\left(F_n + F_{n+1}\right)</math>
<math display=block>\sum_{i=1}^{n+1} F_i^2 = F_{n+1}F_{n+2}</math>
Binet formula proofs
The Binet formula is
<math display=block>\sqrt5F_n = \varphi^n - \psi^n.</math>
This can be used to prove Fibonacci identities.
For example, to prove that <math display=inline>\sum_{i=1}^n F_i = F_{n+2} - 1</math>
note that the left hand side multiplied by <math>\sqrt5</math> becomes
<math display=block>
\begin{align}
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)\\
&= \sqrt5(F_{n+2}-1)\\
\end{align}</math>
as required, using the facts <math display=inline>\varphi\psi =- 1</math> and <math display=inline>\varphi-\psi=\sqrt5</math> to simplify the equations.
Other identities
Numerous other identities can be derived using various methods. Here are some of them:[23]
Cassini's and Catalan's identities
Шаблон:Main
Cassini's identity states that
<math display=block>{F_n}^2 - F_{n+1}F_{n-1} = (-1)^{n-1}</math>
Catalan's identity is a generalization:
<math display=block>{F_n}^2 - F_{n+r}F_{n-r} = (-1)^{n-r}{F_r}^2</math>
d'Ocagne's identity
<math display=block>F_m F_{n+1} - F_{m+1} F_n = (-1)^n F_{m-n}</math>
<math display=block>F_{2 n} = {F_{n+1}}^2 - {F_{n-1}}^2 = F_n \left (F_{n+1}+F_{n-1} \right ) = F_nL_n</math>
where Шаблон:Math is the Шаблон:Mvar-th Lucas number. The last is an identity for doubling Шаблон:Mvar; other identities of this type are
<math display=block>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</math>
by Cassini's identity.
<math display=block>F_{3 n+1} = {F_{n+1}}^3 + 3 F_{n+1}{F_n}^2 - {F_n}^3</math>
<math display=block>F_{3 n+2} = {F_{n+1}}^3 + 3 {F_{n+1}}^2 F_n + {F_n}^3</math>
<math display=block>F_{4 n} = 4 F_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 )</math>
These can be found experimentally using lattice reduction, and are useful in setting up the special number field sieve to factorize a Fibonacci number.
where all terms involving <math>z^k</math> for <math>k \ge 2</math> cancel out because of the defining Fibonacci recurrence relation.
The partial fraction decomposition is given by
<math display=block>s(z) = \frac{1}{\sqrt5}\left(\frac{1}{1 - \varphi z} - \frac{1}{1 - \psi z}\right)</math>
where <math display=inline>\varphi = \tfrac12\left(1 + \sqrt{5}\right)</math> is the golden ratio and <math>\psi = \tfrac12\left(1 - \sqrt{5}\right)</math> is its conjugate.
The related function <math display=inline>z \mapsto -s\left(-1/z\right)</math> is the generating function for the negafibonacci numbers, and <math>s(z)</math> satisfies the functional equation
Using <math>z</math> equal to any of 0.01, 0.001, 0.0001, etc. lays out the first Fibonacci numbers in the decimal expansion of <math>s(z)</math>. For example, <math>s(0.001) = \frac{0.001}{0.998999} = \frac{1000}{998999} = 0.001001002003005008013021\ldots.</math>
Reciprocal sums
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
<math display=block>\sum_{k=1}^\infty \frac{1}{F_{2 k-1}} = \frac{\sqrt{5}}{4} \; \vartheta_2\!\left(0, \frac{3-\sqrt 5}{2}\right)^2 ,</math>
and the sum of squared reciprocal Fibonacci numbers as
<math display=block>\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).</math>
If we add 1 to each Fibonacci number in the first sum, there is also the closed form
<math display=block>\sum_{k=1}^\infty \frac{1}{1+F_{2 k-1}} = \frac{\sqrt{5}}{2},</math>
and there is a nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio,
<math display=block>\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2} .</math>
The sum of all even-indexed reciprocal Fibonacci numbers is[25]
<math display=block>\sum_{k=1}^{\infty} \frac{1}{F_{2 k}} = \sqrt{5} \left(L(\psi^2) - L(\psi^4)\right) </math>
with the Lambert series <math>\textstyle L(q) := \sum_{k=1}^{\infty} \frac{q^k}{1-q^k} ,</math> since <math>\textstyle \frac{1}{F_{2 k}} = \sqrt{5} \left(\frac{\psi^{2 k}}{1-\psi^{2 k}} - \frac{\psi^{4 k}}{1-\psi^{4 k}} \right)\!.</math>
Millin's series gives the identity[28]
<math display=block>\sum_{k=0}^{\infty} \frac{1}{F_{2^k}} = \frac{7 - \sqrt{5}}{2},</math>
which follows from the closed form for its partial sums as Шаблон:Mvar tends to infinity:
<math display=block>\sum_{k=0}^N \frac{1}{F_{2^k}} = 3 - \frac{F_{2^N-1}}{F_{2^N}}.</math>
Primes and divisibility
Divisibility properties
Every third number of the sequence is even (a multiple of <math>F_3=2</math>) and, more generally, every Шаблон:Mvar-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[29][30]
<math display=block>\gcd(F_a,F_b,F_c,\ldots) = F_{\gcd(a,b,c,\ldots)}\,</math>
where Шаблон:Math is the greatest common divisor function.
In particular, any three consecutive Fibonacci numbers are pairwise coprime because both <math>F_1=1</math> and <math>F_2 = 1</math>. That is,
<math display=block>\begin{cases} p =5 & \Rightarrow p \mid F_{p}, \\ p \equiv \pm1 \pmod 5 & \Rightarrow p \mid F_{p-1}, \\ p \equiv \pm2 \pmod 5 & \Rightarrow p \mid F_{p+1}.\end{cases}</math>
These cases can be combined into a single, non-piecewise formula, using the Legendre symbol:[31]
<math display=block>p \mid F_{p \;-\, \left(\frac{5}{p}\right)}.</math>
Primality testing
The above formula can be used as a primality test in the sense that if
<math display=block>n \mid F_{n \;-\, \left(\frac{5}{n}\right)},</math>
where the Legendre symbol has been replaced by the Jacobi symbol, then this is evidence that Шаблон:Mvar is a prime, and if it fails to hold, then Шаблон:Mvar is definitely not a prime. If Шаблон:Mvar is composite and satisfies the formula, then Шаблон:Mvar is a Fibonacci pseudoprime. When Шаблон:Mvar is largeШаблон:Sndsay a 500-bit numberШаблон:Sndthen we can calculate Шаблон:Math efficiently using the matrix form. Thus
<math display=block> \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.</math>
Here the matrix power Шаблон:Math is calculated using modular exponentiation, which can be adapted to matrices.[32]
No Fibonacci number greater than Шаблон:Math is one greater or one less than a prime number.[35]
The only nontrivial square Fibonacci number is 144.[36] Attila Pethő proved in 2001 that there is only a finite number of perfect power Fibonacci numbers.[37] In 2006, Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only such non-trivial perfect powers.[38]
No Fibonacci number can be a perfect number.[40] More generally, no Fibonacci number other than 1 can be multiply perfect,[41] and no ratio of two Fibonacci numbers can be perfect.[42]
Prime divisors
With the exceptions of 1, 8 and 144 (Шаблон:Math, Шаблон:Math and Шаблон:Math) every Fibonacci number has a prime factor that is not a factor of any smaller Fibonacci number (Carmichael's theorem).[43] As a result, 8 and 144 (Шаблон:Math and Шаблон:Math) are the only Fibonacci numbers that are the product of other Fibonacci numbers.[44]
The divisibility of Fibonacci numbers by a prime Шаблон:Mvar is related to the Legendre symbol <math>\left(\tfrac{p}{5}\right)</math> which is evaluated as follows:
<math display=block>\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}</math>
If Шаблон:Mvar is a prime number then
<math display=block> 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.</math>[45]Шаблон:Sfn
Also, if Шаблон:Math is an odd prime number then:Шаблон:Sfn
<math display=block>5 {F_{\frac{p \pm 1}{2}}}^2 \equiv \begin{cases}
\tfrac{1}{2} \left (5\left(\frac{p}{5}\right)\pm 5 \right ) \pmod p & \text{if } p \equiv 1 \pmod 4\\
\tfrac{1}{2} \left (5\left(\frac{p}{5}\right)\mp 3 \right ) \pmod p & \text{if } p \equiv 3 \pmod 4.
\end{cases}</math>
Example 1.Шаблон:Math, in this case Шаблон:Math and we have:
<math display=block>(\tfrac{7}{5}) = -1: \qquad \tfrac{1}{2}\left (5(\tfrac{7}{5})+3 \right ) =-1, \quad \tfrac{1}{2} \left (5(\tfrac{7}{5})-3 \right )=-4.</math>
<math display=block>F_3=2 \text{ and } F_4=3.</math>
<math display=block>5{F_3}^2=20\equiv -1 \pmod {7}\;\;\text{ and }\;\;5{F_4}^2=45\equiv -4 \pmod {7}</math>
Example 2.Шаблон:Math, in this case Шаблон:Math and we have:
<math display=block>(\tfrac{11}{5}) = +1: \qquad \tfrac{1}{2}\left (5(\tfrac{11}{5})+3 \right )=4, \quad \tfrac{1}{2} \left (5(\tfrac{11}{5})- 3 \right )=1.</math>
<math display=block>F_5=5 \text{ and } F_6=8.</math>
<math display=block>5{F_5}^2=125\equiv 4 \pmod {11} \;\;\text{ and }\;\;5{F_6}^2=320\equiv 1 \pmod {11}</math>
Example 3.Шаблон:Math, in this case Шаблон:Math and we have:
<math display=block>(\tfrac{13}{5}) = -1: \qquad \tfrac{1}{2}\left (5(\tfrac{13}{5})-5 \right ) =-5, \quad \tfrac{1}{2}\left (5(\tfrac{13}{5})+ 5 \right )=0.</math>
<math display=block>F_6=8 \text{ and } F_7=13.</math>
<math display=block>5{F_6}^2=320\equiv -5 \pmod {13} \;\;\text{ and }\;\;5{F_7}^2=845\equiv 0 \pmod {13}</math>
Example 4.Шаблон:Math, in this case Шаблон:Math and we have:
<math display=block>(\tfrac{29}{5}) = +1: \qquad \tfrac{1}{2}\left (5(\tfrac{29}{5})-5 \right )=0, \quad \tfrac{1}{2}\left (5(\tfrac{29}{5})+5 \right )=5.</math>
<math display=block>F_{14}=377 \text{ and } F_{15}=610.</math>
<math display=block>5{F_{14}}^2=710645\equiv 0 \pmod {29} \;\;\text{ and }\;\;5{F_{15}}^2=1860500\equiv 5 \pmod {29}</math>
For odd Шаблон:Mvar, all odd prime divisors of Шаблон:Math are congruent to 1 modulo 4, implying that all odd divisors of Шаблон:Math (as the products of odd prime divisors) are congruent to 1 modulo 4.Шаблон:Sfn
Letting a number be a linear function (other than the sum) of the 2 preceding numbers. The Pell numbers have Шаблон:Math. If the coefficient of the preceding value is assigned a variable value Шаблон:Mvar, the result is the sequence of Fibonacci polynomials.
Generating the next number by adding 3 numbers (tribonacci numbers), 4 numbers (tetranacci numbers), or more. The resulting sequences are known as n-Step Fibonacci numbers.[50]
The Fibonacci numbers occur as the sums of binomial coefficients in the "shallow" diagonals of Pascal's triangle:Шаблон:Sfn
<math display=block>F_n = \sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-k-1}{k}.</math>
This can be proved by expanding the generating function
<math display=block>\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</math>
and collecting like terms of <math>x^n</math>.
To see how the formula is used, we can arrange the sums by the number of terms present:
These numbers also give the solution to certain enumerative problems,[51] the most common of which is that of counting the number of ways of writing a given number Шаблон:Mvar as an ordered sum of 1s and 2s (called compositions); there are Шаблон:Math ways to do this (equivalently, it's also the number of domino tilings of the <math>2\times n</math> rectangle). For example, there are Шаблон:Math 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 binarystrings, or equivalently, among the subsets of a given set.
The number of binary strings of length Шаблон:Mvar without consecutive Шаблон:Maths is the Fibonacci number Шаблон:Math. For example, out of the 16 binary strings of length 4, there are Шаблон:Math without consecutive Шаблон:Maths—they are 0000, 0001, 0010, 0100, 0101, 1000, 1001, and 1010. Such strings are the binary representations of Fibbinary numbers. Equivalently, Шаблон:Math is the number of subsets Шаблон:Mvar of Шаблон:Math without consecutive integers, that is, those Шаблон:Mvar for which Шаблон:Math for every Шаблон:Mvar. A bijection with the sums to Шаблон:Math is to replace 1 with 0 and 2 with 10, and drop the last zero.
The number of binary strings of length Шаблон:Mvar without an odd number of consecutive Шаблон:Maths is the Fibonacci number Шаблон:Math. For example, out of the 16 binary strings of length 4, there are Шаблон:Math without an odd number of consecutive Шаблон:Maths—they are 0000, 0011, 0110, 1100, 1111. Equivalently, the number of subsets Шаблон:Mvar of Шаблон:Math without an odd number of consecutive integers is Шаблон:Math. A bijection with the sums to Шаблон:Mvar is to replace 1 with 0 and 2 with 11.
The number of binary strings of length Шаблон:Mvar without an even number of consecutive Шаблон:Maths or Шаблон:Maths is Шаблон:Math. For example, out of the 16 binary strings of length 4, there are Шаблон:Math without an even number of consecutive Шаблон:Maths or Шаблон:Maths—they are 0001, 0111, 0101, 1000, 1010, 1110. There is an equivalent statement about subsets.
The Fibonacci numbers are also an example of a complete 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 of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. This is known as Zeckendorf's theorem, and a sum of Fibonacci numbers that satisfies these conditions is called a Zeckendorf representation. The Zeckendorf representation of a number can be used to derive its Fibonacci coding.
Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple, obtained from the formula <math display=block>(F_n F_{n+3})^2 + (2 F_{n+1}F_{n+2})^2 = {F_{2 n+3}}^2.</math> 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.[53]
Fibonacci numbers are used in a polyphase version of the merge sort algorithm in which an unsorted list is divided into two lists whose lengths correspond to sequential Fibonacci numbers—by dividing the list so that the two parts have lengths in the approximate proportion Шаблон:Mvar. A tape-drive implementation of the polyphase merge sort was described in The Art of Computer Programming.
Шаблон:AnchorA Fibonacci tree is a binary tree whose child trees (recursively) differ in height by exactly 1. So it is an AVL tree, and one with the fewest nodes for a given height—the "thinnest" AVL tree. These trees have a number of vertices that is a Fibonacci number minus one, an important fact in the analysis of AVL trees.[56]
The Fibonacci number series is used for optional lossy compression in the IFF8SVX audio file format used on Amiga computers. The number series compands the original audio wave similar to logarithmic methods such as μ-law.[58][59]
Some Agile teams use a modified series called the "Modified Fibonacci Series" in planning poker, as an estimation tool. Planning Poker is a formal part of the Scaled Agile Framework.[60]
Файл:FibonacciChamomile.PNGYellow 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.
<math display=block>\theta = \frac{2\pi}{\varphi^2} n,\ r = c \sqrt{n}</math>
where Шаблон:Mvar is the index number of the floret and Шаблон:Mvar 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 Шаблон:Math, the nearest neighbors of floret number Шаблон:Mvar are those at Шаблон:Math for some index Шаблон:Mvar, which depends on Шаблон:Mvar, 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,Шаблон:Sfn typically counted by the outermost range of radii.[69]
Fibonacci numbers also appear in the pedigrees of idealized honeybees, according to the following rules:
If an egg is laid by an unmated female, it hatches a male or drone bee.
If, however, an egg was fertilized by a male, it hatches 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, Шаблон:Math, is the number of female ancestors, which is Шаблон:Math, plus the number of male ancestors, which is Шаблон:Math.[70] This is under the unrealistic assumption that the ancestors at each level are otherwise unrelated.
Файл:X chromosome ancestral line Fibonacci sequence.svgThe 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".[71])
It has 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.[71] 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 (<math>F_1=1</math>), and at his parents' generation, his X chromosome came from a single parent Шаблон:Nowrap. 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 Шаблон:Nowrap. 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 Шаблон:Nowrap. Five great-great-grandparents contributed to the male descendant's X chromosome Шаблон:Nowrap, 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.)
Other
In optics, when a beam of light shines at an angle through two stacked transparent plates of different materials of different refractive indexes, it may reflect off three surfaces: the top, middle, and bottom surfaces of the two plates. The number of different beam paths that have Шаблон:Mvar reflections, for Шаблон:Math, is the Шаблон:Mvar-th Fibonacci number. (However, when Шаблон:Math, there are three reflection paths, not two, one for each of the three surfaces.)Шаблон:Sfn
Since the conversion 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 a radix 2 number register in golden ratio baseШаблон:Mvar being shifted. To convert from kilometers to miles, shift the register down the Fibonacci sequence instead.[72]
The measured values of voltages and currents in the infinite resistor chain circuit (also called the resistor 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.[73]
Brasch et al. 2012 show how a generalized Fibonacci sequence also can be connected to the field of economics.[74] 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.Шаблон:Sfn
