Fibonacci number

Joseph Schillinger (1895–1943) developed a 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.[31]

Yellow chamomile head showing the arrangement in 21 (blue) and 13 (aqua) spirals. Such arrangements involving consecutive Fibonacci numbers appear in a wide variety of plants.

Fibonacci sequences appear in biological settings,[32] such as branching in trees, arrangement of leaves on a stem, the fruitlets of a pineapple,[33] the flowering of artichoke, an uncurling fern and the arrangement of a pine cone,[34] and the family tree of honeybees.[35][36] Kepler pointed out the presence of the Fibonacci sequence in nature, using it to explain the (golden ratio-related) pentagonal form of some flowers.[37] Field daisies most often have petals in counts of Fibonacci numbers.[38] In 1754, Charles Bonnet discovered that the spiral phyllotaxis of plants were frequently expressed in Fibonacci number series.[39]

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.[40]

Illustration of Vogel's model for n = 1 ... 500

A model for the pattern of florets in the head of a sunflower was proposed by Helmut Vogel in 1979.[41] This has the form

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

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,[42] typically counted by the outermost range of radii.[43]

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, F_n, is the number of female ancestors, which is F_n−1, plus the number of male ancestors, which is F_n−2.[44] 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".[45])

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.[45] 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 (${\displaystyle F_{1}=1}$), and at his parents' generation, his X chromosome came from a single parent (${\displaystyle F_{2}=1}$). 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 (${\displaystyle F_{3}=2}$). 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 (${\displaystyle F_{4}=3}$). Five great-great-grandparents contributed to the male descendant's X chromosome (${\displaystyle F_{5}=5}$), 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.)

The pathways of tubulins on intracellular microtubules arrange in patterns of 3, 5, 8 and 13.[46]

The first 21 Fibonacci numbers F_n are:[2]

F0	F1	F2	F3	F4	F5	F6	F7	F8	F9	F10	F11	F12	F13	F14	F15	F16	F17	F18	F19	F20
0	1	1	2	3	5	8	13	21	34	55	89	144	233	377	610	987	1597	2584	4181	6765

The sequence can also be extended to negative index n using the re-arranged recurrence relation

${\displaystyle F_{n-2}=F_{n}-F_{n-1},}$

which yields the sequence of "negafibonacci" numbers[49] satisfying

${\displaystyle F_{-n}=(-1)^{n+1}F_{n}.}$

Thus the bidirectional sequence is

F−8

F−7

F−6

F−5

F−4

F−3

F−2

F−1

F0

F1

F2

F3

F4

F5

F6

F7

F8

−21

13

−8

5

−3

2

−1

1

0

1

1

2

3

5

8

13

21

Like every sequence defined by a linear recurrence with constant coefficients, the Fibonacci numbers have a closed-form solution. It has become known as "Binet's formula", though it was already known by Abraham de Moivre and Daniel Bernoulli:[50]

${\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}={\frac {\varphi ^{n}-\psi ^{n}}{\sqrt {5}}}}$

where

${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}\approx 1.61803\,39887\ldots }$

is the golden ratio (OEIS: A001622), and

${\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}=1-\varphi =-{1 \over \varphi }\approx -0.61803\,39887\ldots .}$[51]

Since ${\displaystyle \psi =-\varphi ^{-1}}$, this formula can also be written as

${\displaystyle F_{n}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}={\frac {\varphi ^{n}-(-\varphi )^{-n}}{2\varphi -1}}}$

To see this,[52] note that φ and ψ are both solutions of the equations

${\displaystyle x^{2}=x+1\quad {\text{and}}\quad x^{n}=x^{n-1}+x^{n-2},}$

so the powers of φ and ψ satisfy the Fibonacci recursion. In other words,

${\displaystyle \varphi ^{n}=\varphi ^{n-1}+\varphi ^{n-2}}$

and

${\displaystyle \psi ^{n}=\psi ^{n-1}+\psi ^{n-2}.}$

It follows that for any values a and b, the sequence defined by

${\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}$

satisfies the same recurrence

${\displaystyle U_{n}=a\varphi ^{n-1}+b\psi ^{n-1}+a\varphi ^{n-2}+b\psi ^{n-2}=U_{n-1}+U_{n-2}.}$

If a and b are chosen so that U₀ = 0 and U₁ = 1 then the resulting sequence U_n must be the Fibonacci sequence. This is the same as requiring a and b satisfy the system of equations:

${\displaystyle \left\{{\begin{array}{l}a+b=0\\\varphi a+\psi b=1\end{array}}\right.}$

which has solution

${\displaystyle a={\frac {1}{\varphi -\psi }}={\frac {1}{\sqrt {5}}},\quad b=-a,}$

producing the required formula.

Taking the starting values U₀ and U₁ to be arbitrary constants, a more general solution is:

${\displaystyle U_{n}=a\varphi ^{n}+b\psi ^{n}}$

where

${\displaystyle a={\frac {U_{1}-U_{0}\psi }{\sqrt {5}}}}$

${\displaystyle b={\frac {U_{0}\varphi -U_{1}}{\sqrt {5}}}}$.

Since

${\displaystyle \left|{\frac {\psi ^{n}}{\sqrt {5}}}\right|<{\frac {1}{2}}}$

for all n ≥ 0, the number F_n is the closest integer to ${\displaystyle {\frac {\varphi ^{n}}{\sqrt {5}}}}$. Therefore, it can be found by rounding, using the nearest integer function:

${\displaystyle F_{n}=\left[{\frac {\varphi ^{n}}{\sqrt {5}}}\right],\ n\geq 0.}$

In fact, the rounding error is very small, being less than 0.1 for n ≥ 4, and less than 0.01 for n ≥ 8.

Fibonacci number can also be computed by truncation, in terms of the floor function:

${\displaystyle F_{n}=\left\lfloor {\frac {\varphi ^{n}}{\sqrt {5}}}+{\frac {1}{2}}\right\rfloor ,\ n\geq 0.}$

As the floor function is monotonic, the latter formula can be inverted for finding the index n(F) of the largest Fibonacci number that is not greater than a real number F > 1:

${\displaystyle n(F)=\left\lfloor \log _{\varphi }\left(F\cdot {\sqrt {5}}+{\frac {1}{2}}\right)\right\rfloor ,}$

where ${\displaystyle \log _{\varphi }(x)=\ln(x)/\ln(\varphi )=\log _{10}(x)/\log _{10}(\varphi ).}$

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 ${\displaystyle \varphi \colon }$[53][54]

${\displaystyle \lim _{n\to \infty }{\frac {F_{n+1}}{F_{n}}}=\varphi .}$

This convergence holds regardless of the starting values, excluding 0 and 0, or any pair in the conjugate golden ratio, ${\displaystyle -1/\varphi .}$ 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 terms in this sequence shows the same convergence towards the golden ratio.

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

Since the golden ratio satisfies the equation

${\displaystyle \varphi ^{2}=\varphi +1,}$

this expression can be used to decompose higher powers ${\displaystyle \varphi ^{n}}$ as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of ${\displaystyle \varphi }$ and 1. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients:

${\displaystyle \varphi ^{n}=F_{n}\varphi +F_{n-1}.}$

This equation can be proved by induction on n.

This expression is also true for n < 1 if the Fibonacci sequence F_n is extended to negative integers using the Fibonacci rule ${\displaystyle F_{n}=F_{n-1}+F_{n-2}.}$

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

${\displaystyle {F_{k+2} \choose F_{k+1}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{F_{k+1} \choose F_{k}}}$

alternatively denoted

${\displaystyle {\vec {F}}_{k+1}=\mathbf {A} {\vec {F}}_{k},}$

which yields ${\displaystyle {\vec {F}}_{n}=\mathbf {A} ^{n}{\vec {F}}_{0}}$. The eigenvalues of the matrix A are ${\displaystyle \varphi ={\frac {1}{2}}(1+{\sqrt {5}})}$ and ${\displaystyle -\varphi ^{-1}={\frac {1}{2}}(1-{\sqrt {5}})}$ corresponding to the respective eigenvectors

${\displaystyle {\vec {\mu }}={\varphi \choose 1}}$

and

${\displaystyle {\vec {\nu }}={-\varphi ^{-1} \choose 1}.}$

As the initial value is

${\displaystyle {\vec {F}}_{0}={1 \choose 0}={\frac {1}{\sqrt {5}}}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}{\vec {\nu }},}$

it follows that the nth term is

${\displaystyle {\begin{aligned}{\vec {F}}_{n}&={\frac {1}{\sqrt {5}}}A^{n}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}A^{n}{\vec {\nu }}\\&={\frac {1}{\sqrt {5}}}\varphi ^{n}{\vec {\mu }}-{\frac {1}{\sqrt {5}}}(-\varphi )^{-n}{\vec {\nu }}~\\&={\cfrac {1}{\sqrt {5}}}\left({\cfrac {1+{\sqrt {5}}}{2}}\right)^{n}{\varphi \choose 1}-{\cfrac {1}{\sqrt {5}}}\left({\cfrac {1-{\sqrt {5}}}{2}}\right)^{n}{-\varphi ^{-1} \choose 1},\end{aligned}}}$

From this, the nth element in the Fibonacci series may be read off directly as a closed-form expression:

${\displaystyle F_{n}={\cfrac {1}{\sqrt {5}}}\left({\cfrac {1+{\sqrt {5}}}{2}}\right)^{n}-{\cfrac {1}{\sqrt {5}}}\left({\cfrac {1-{\sqrt {5}}}{2}}\right)^{n}.}$

Equivalently, the same computation may performed by diagonalization of A through use of its eigendecomposition:

${\displaystyle {\begin{aligned}A&=S\Lambda S^{-1},\\A^{n}&=S\Lambda ^{n}S^{-1},\end{aligned}}}$

where ${\displaystyle \Lambda ={\begin{pmatrix}\varphi &0\\0&-\varphi ^{-1}\end{pmatrix}}}$ and ${\displaystyle S={\begin{pmatrix}\varphi &-\varphi ^{-1}\\1&1\end{pmatrix}}.}$ The closed-form expression for the nth element in the Fibonacci series is therefore given by

${\displaystyle {\begin{aligned}{F_{n+1} \choose F_{n}}&=A^{n}{F_{1} \choose F_{0}}\\&=S\Lambda ^{n}S^{-1}{F_{1} \choose F_{0}}\\&=S{\begin{pmatrix}\varphi ^{n}&0\\0&(-\varphi )^{-n}\end{pmatrix}}S^{-1}{F_{1} \choose F_{0}}\\&={\begin{pmatrix}\varphi &-\varphi ^{-1}\\1&1\end{pmatrix}}{\begin{pmatrix}\varphi ^{n}&0\\0&(-\varphi )^{-n}\end{pmatrix}}{\frac {1}{\sqrt {5}}}{\begin{pmatrix}1&\varphi ^{-1}\\-1&\varphi \end{pmatrix}}{1 \choose 0},\end{aligned}}}$

which again yields

${\displaystyle F_{n}={\cfrac {\varphi ^{n}-(-\varphi )^{-n}}{\sqrt {5}}}.}$

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:

${\displaystyle \varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}}}.}$

The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for φ, 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:

${\displaystyle {\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n}={\begin{pmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{pmatrix}}.}$

Taking the determinant of both sides of this equation yields Cassini's identity,

${\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-F_{n}^{2}.}$

Moreover, since Aⁿ A^m = A^n+m 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),

${\displaystyle {\begin{aligned}{F_{m}}{F_{n}}+{F_{m-1}}{F_{n-1}}&=F_{m+n-1},\\F_{m}F_{n+1}+F_{m-1}F_{n}&=F_{m+n}.\end{aligned}}}$

In particular, with m = n,

${\displaystyle {\begin{aligned}F_{2n-1}&=F_{n}^{2}+F_{n-1}^{2}\\F_{2n}&=(F_{n-1}+F_{n+1})F_{n}\\&=(2F_{n-1}+F_{n})F_{n}.\end{aligned}}}$

These last two identities provide a way to compute Fibonacci numbers recursively in O(log(n)) arithmetic operations and in time O(M(n) log(n)), where M(n) is the time for the multiplication of two numbers of n digits. This matches the time for computing the nth 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).[55]

The question may arise whether a positive integer x is a Fibonacci number. This is true if and only if at least one of ${\displaystyle 5x^{2}+4}$ or ${\displaystyle 5x^{2}-4}$ is a perfect square.[56] This is because Binet's formula above can be rearranged to give

${\displaystyle n=\log _{\varphi }\left({\frac {F_{n}{\sqrt {5}}+{\sqrt {5F_{n}^{2}\pm 4}}}{2}}\right),}$

which allows one to find the position in the sequence of a given Fibonacci number.

This formula must return an integer for all n, so the radical expression must be an integer (otherwise the logarithm does not even return a rational number).

Most identities involving Fibonacci numbers can be proved using combinatorial arguments using the fact that F_n can be interpreted as the number of sequences of 1s and 2s that sum to n − 1. This can be taken as the definition of F_n, with the convention that F₀ = 0, meaning no sum adds up to −1, and that F₁ = 1, meaning the empty sum "adds up" to 0. Here, the order of the summand matters. For example, 1 + 2 and 2 + 1 are considered two different sums.

For example, the recurrence relation

${\displaystyle F_{n}=F_{n-1}+F_{n-2},}$

or in words, the nth Fibonacci number is the sum of the previous two Fibonacci numbers, may be shown by dividing the F_n sums of 1s and 2s that add to n − 1 into two non-overlapping groups. One group contains those sums whose first term is 1 and the other those sums whose first term is 2. In the first group the remaining terms add to n − 2, so it has F_n-1 sums, and in the second group the remaining terms add to n − 3, so there are F_n−2 sums. So there are a total of F_n−1 + F_n−2 sums altogether, showing this is equal to F_n.

Similarly, it may be shown that the sum of the first Fibonacci numbers up to the nth is equal to the (n + 2)-nd Fibonacci number minus 1.[57] In symbols:

${\displaystyle \sum _{i=1}^{n}F_{i}=F_{n+2}-1}$

This is done by dividing the sums adding to n + 1 in a different way, this time by the location of the first 2. Specifically, the first group consists of those sums that start with 2, the second group those that start 1 + 2, the third 1 + 1 + 2, and so on, until the last group, which consists of the single sum where only 1's are used. The number of sums in the first group is F(n), F(n − 1) in the second group, and so on, with 1 sum in the last group. So the total number of sums is F(n) + F(n − 1) + ... + F(1) + 1 and therefore this quantity is equal to F(n + 2).

A similar argument, grouping the sums by the position of the first 1 rather than the first 2, gives two more identities:

${\displaystyle \sum _{i=0}^{n-1}F_{2i+1}=F_{2n}}$

and

${\displaystyle \sum _{i=1}^{n}F_{2i}=F_{2n+1}-1.}$

In words, the sum of the first Fibonacci numbers with odd index up to F_2n−1 is the (2n)th Fibonacci number, and the sum of the first Fibonacci numbers with even index up to F_2n is the (2n + 1)th Fibonacci number minus 1.[58]

A different trick may be used to prove

${\displaystyle \sum _{i=1}^{n}{F_{i}}^{2}=F_{n}F_{n+1},}$

or in words, the sum of the squares of the first Fibonacci numbers up to F_n is the product of the nth and (n + 1)th Fibonacci numbers. In this case Fibonacci rectangle of size F_n by F(n + 1) can be decomposed into squares of size F_n, F_n−1, and so on to F₁ = 1, from which the identity follows by comparing areas.

The sequence ${\displaystyle (F_{n})_{n\in \mathbb {N} }}$ is also considered using the symbolic method.[59] More precisely, this sequence corresponds to a specifiable combinatorial class. The specification of this sequence is ${\displaystyle \operatorname {Seq} ({\mathcal {Z+Z^{2}}})}$. Indeed, as stated above, the ${\displaystyle n}$-th Fibonacci number equals the number of combinatorial compositions (ordered partitions) of ${\displaystyle n-1}$ using terms 1 and 2.

It follows that the ordinary generating function of the Fibonacci sequence, i.e. ${\displaystyle \sum _{i=0}^{\infty }F_{i}z^{i}}$, is the complex function ${\displaystyle {\frac {z}{1-z-z^{2}}}}$.

Numerous other identities can be derived using various methods. Some of the most noteworthy are:[60]

Cassini's identity states that

${\displaystyle F_{n}^{2}-F_{n+1}F_{n-1}=(-1)^{n-1}}$

Catalan's identity is a generalization:

${\displaystyle F_{n}^{2}-F_{n+r}F_{n-r}=(-1)^{n-r}F_{r}^{2}}$

${\displaystyle F_{m}F_{n+1}-F_{m+1}F_{n}=(-1)^{n}F_{m-n}}$

${\displaystyle F_{2n}=F_{n+1}^{2}-F_{n-1}^{2}=F_{n}\left(F_{n+1}+F_{n-1}\right)=F_{n}L_{n}}$

where L_n is the n'th Lucas number. The last is an identity for doubling n; other identities of this type are

${\displaystyle F_{3n}=2F_{n}^{3}+3F_{n}F_{n+1}F_{n-1}=5F_{n}^{3}+3(-1)^{n}F_{n}}$

by Cassini's identity.

${\displaystyle F_{3n+1}=F_{n+1}^{3}+3F_{n+1}F_{n}^{2}-F_{n}^{3}}$

${\displaystyle F_{3n+2}=F_{n+1}^{3}+3F_{n+1}^{2}F_{n}+F_{n}^{3}}$

${\displaystyle F_{4n}=4F_{n}F_{n+1}\left(F_{n+1}^{2}+2F_{n}^{2}\right)-3F_{n}^{2}\left(F_{n}^{2}+2F_{n+1}^{2}\right)}$

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,[60]

${\displaystyle F_{kn+c}=\sum _{i=0}^{k}{k \choose i}F_{c-i}F_{n}^{i}F_{n+1}^{k-i}.}$

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

${\displaystyle s(x)=\sum _{k=0}^{\infty }F_{k}x^{k}.}$

This series is convergent for ${\displaystyle |x|<{\frac {1}{\varphi }},}$ and its sum has a simple closed-form:[61]

${\displaystyle s(x)={\frac {x}{1-x-x^{2}}}}$

This can be proved by using the Fibonacci recurrence to expand each coefficient in the infinite sum:

${\displaystyle {\begin{aligned}s(x)&=\sum _{k=0}^{\infty }F_{k}x^{k}\\&=F_{0}+F_{1}x+\sum _{k=2}^{\infty }\left(F_{k-1}+F_{k-2}\right)x^{k}\\&=x+\sum _{k=2}^{\infty }F_{k-1}x^{k}+\sum _{k=2}^{\infty }F_{k-2}x^{k}\\&=x+x\sum _{k=0}^{\infty }F_{k}x^{k}+x^{2}\sum _{k=0}^{\infty }F_{k}x^{k}\\&=x+xs(x)+x^{2}s(x).\end{aligned}}}$

Solving the equation

${\displaystyle s(x)=x+xs(x)+x^{2}s(x)}$

for s(x) results in the above closed form.

Setting x = 1/k, the closed form of the series becomes

${\displaystyle \sum _{n=0}^{\infty }{\frac {F_{n}}{k^{n}}}={\frac {k}{k^{2}-k-1}}.}$

In particular, if k is an integer greater than 1, then this series converges. Further setting k = 10^m yields

${\displaystyle \sum _{n=1}^{\infty }{\frac {F_{n}}{10^{m(n+1)}}}={\frac {1}{10^{2m}-10^{m}-1}}}$

for all positive integers m.

Some math puzzle-books present as curious the particular value that comes from m = 1, which is ${\displaystyle {\frac {s(1/10)}{10}}={\frac {1}{89}}=.011235\ldots }$[62] Similarly, m = 2 gives

${\displaystyle {\frac {s(1/100)}{100}}={\frac {1}{9899}}=.0001010203050813213455\ldots }$

Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of theta functions. For example, we can write the sum of every odd-indexed reciprocal Fibonacci number as

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{F_{2k+1}}}={\frac {\sqrt {5}}{4}}\vartheta _{2}^{2}\left(0,{\frac {3-{\sqrt {5}}}{2}}\right),}$

and the sum of squared reciprocal Fibonacci numbers as

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{F_{k}^{2}}}={\frac {5}{24}}\left(\vartheta _{2}^{4}\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)-\vartheta _{4}^{4}\left(0,{\frac {3-{\sqrt {5}}}{2}}\right)+1\right).}$

If we add 1 to each Fibonacci number in the first sum, there is also the closed form

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{1+F_{2k+1}}}={\frac {\sqrt {5}}{2}},}$

and there is a nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio,

${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{\sum _{j=1}^{k}{F_{j}}^{2}}}={\frac {{\sqrt {5}}-1}{2}}.}$

No closed formula for the reciprocal Fibonacci constant

${\displaystyle \psi =\sum _{k=1}^{\infty }{\frac {1}{F_{k}}}=3.359885666243\dots }$

is known, but the number has been proved irrational by Richard André-Jeannin.[63]

The Millin series gives the identity[64]

${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{F_{2^{n}}}}={\frac {7-{\sqrt {5}}}{2}},}$

which follows from the closed form for its partial sums as N tends to infinity:

${\displaystyle \sum _{n=0}^{N}{\frac {1}{F_{2^{n}}}}=3-{\frac {F_{2^{N}-1}}{F_{2^{N}}}}.}$

Every third number of the sequence is even and more generally, every kth number of the sequence is a multiple of F_k. Thus the Fibonacci sequence is an example of a divisibility sequence. In fact, the Fibonacci sequence satisfies the stronger divisibility property[65][66]

${\displaystyle \gcd(F_{m},F_{n})=F_{\gcd(m,n)}.}$

Any three consecutive Fibonacci numbers are pairwise coprime, which means that, for every n,

gcd(F_n, F_n+1) = gcd(F_n, F_n+2) = gcd(F_n+1, F_n+2) = 1.

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 (mod 5), then p divides F_p − 1, and if p is congruent to 2 or 3 (mod 5), then, p divides F_p + 1. The remaining case is that p = 5, and in this case p divides F_p.

${\displaystyle {\begin{cases}p=5&\Rightarrow p\mid F_{p},\\p\equiv \pm 1{\pmod {5}}&\Rightarrow p\mid F_{p-1},\\p\equiv \pm 2{\pmod {5}}&\Rightarrow p\mid F_{p+1}.\end{cases}}}$

These cases can be combined into a single, non-piecewise formula, using the Legendre symbol:[67]

${\displaystyle p\mid F_{p-\left({\frac {5}{p}}\right)}.}$

The above formula can be used as a primality test in the sense that if

${\displaystyle n\mid F_{n-\left({\frac {5}{n}}\right)},}$

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 F_m (mod n) efficiently using the matrix form. Thus

${\displaystyle {\begin{pmatrix}F_{m+1}&F_{m}\\F_{m}&F_{m-1}\end{pmatrix}}\equiv {\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{m}{\pmod {n}}.}$

Here the matrix power A^m is calculated using modular exponentiation, which can be adapted to matrices.[68]

A Fibonacci prime is a Fibonacci number that is prime. The first few are:

2, 3, 5, 13, 89, 233, 1597, 28657, 514229, ... OEIS: A005478.

Fibonacci primes with thousands of digits have been found, but it is not known whether there are infinitely many.[69]

F_kn is divisible by F_n, 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.[70]

The only nontrivial square Fibonacci number is 144.[71] Attila Pethő proved in 2001 that there is only a finite number of perfect power Fibonacci numbers.[72] In 2006, Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only such non-trivial perfect powers.[73]

1, 3, 21, 55 are the only triangular Fibonacci numbers, which was conjectured by Vern Hoggatt and proved by Luo Ming.[74]

No Fibonacci number can be a perfect number.[75] More generally, no Fibonaci number other than 1 can be multiply perfect,[76] and no ratio of two Fibonacci numbers can be perfect.[77]

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).[78] As a result, 8 and 144 (F₆ and F₁₂) are the only Fibonacci numbers that are the product of other Fibonacci numbers OEIS: A235383.

The divisibility of Fibonacci numbers by a prime p is related to the Legendre symbol ${\displaystyle \left({\tfrac {p}{5}}\right)}$ which is evaluated as follows:

${\displaystyle \left({\frac {p}{5}}\right)={\begin{cases}0&{\text{if }}p=5\\1&{\text{if }}p\equiv \pm 1{\pmod {5}}\\-1&{\text{if }}p\equiv \pm 2{\pmod {5}}.\end{cases}}}$

If p is a prime number then

${\displaystyle F_{p}\equiv \left({\frac {p}{5}}\right){\pmod {p}}\quad {\text{and}}\quad F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p}}.}$[79][80]

For example,

${\displaystyle {\begin{aligned}({\tfrac {2}{5}})&=-1,&F_{3}&=2,&F_{2}&=1,\\({\tfrac {3}{5}})&=-1,&F_{4}&=3,&F_{3}&=2,\\({\tfrac {5}{5}})&=0,&F_{5}&=5,\\({\tfrac {7}{5}})&=-1,&F_{8}&=21,&F_{7}&=13,\\({\tfrac {11}{5}})&=+1,&F_{10}&=55,&F_{11}&=89.\end{aligned}}}$

It is not known whether there exists a prime p such that

${\displaystyle F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p^{2}}}.}$

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

Also, if p ≠ 5 is an odd prime number then:[81]

${\displaystyle 5F_{\frac {p\pm 1}{2}}^{2}\equiv {\begin{cases}{\tfrac {1}{2}}\left(5\left({\frac {p}{5}}\right)\pm 5\right){\pmod {p}}&{\text{if }}p\equiv 1{\pmod {4}}\\{\tfrac {1}{2}}\left(5\left({\frac {p}{5}}\right)\mp 3\right){\pmod {p}}&{\text{if }}p\equiv 3{\pmod {4}}.\end{cases}}}$

Example 1. p = 7, in this case p ≡ 3 (mod 4) and we have:

${\displaystyle ({\tfrac {7}{5}})=-1:\qquad {\tfrac {1}{2}}\left(5({\tfrac {7}{5}})+3\right)=-1,\quad {\tfrac {1}{2}}\left(5({\tfrac {7}{5}})-3\right)=-4.}$

${\displaystyle F_{3}=2{\text{ and }}F_{4}=3.}$

${\displaystyle 5F_{3}^{2}=20\equiv -1{\pmod {7}}\;\;{\text{ and }}\;\;5F_{4}^{2}=45\equiv -4{\pmod {7}}}$

Example 2. p = 11, in this case p ≡ 3 (mod 4) and we have:

${\displaystyle ({\tfrac {11}{5}})=+1:\qquad {\tfrac {1}{2}}\left(5({\tfrac {11}{5}})+3\right)=4,\quad {\tfrac {1}{2}}\left(5({\tfrac {11}{5}})-3\right)=1.}$

${\displaystyle F_{5}=5{\text{ and }}F_{6}=8.}$

${\displaystyle 5F_{5}^{2}=125\equiv 4{\pmod {11}}\;\;{\text{ and }}\;\;5F_{6}^{2}=320\equiv 1{\pmod {11}}}$

Example 3. p = 13, in this case p ≡ 1 (mod 4) and we have:

${\displaystyle ({\tfrac {13}{5}})=-1:\qquad {\tfrac {1}{2}}\left(5({\tfrac {13}{5}})-5\right)=-5,\quad {\tfrac {1}{2}}\left(5({\tfrac {13}{5}})+5\right)=0.}$

${\displaystyle F_{6}=8{\text{ and }}F_{7}=13.}$

${\displaystyle 5F_{6}^{2}=320\equiv -5{\pmod {13}}\;\;{\text{ and }}\;\;5F_{7}^{2}=845\equiv 0{\pmod {13}}}$

Example 4. p = 29, in this case p ≡ 1 (mod 4) and we have:

${\displaystyle ({\tfrac {29}{5}})=+1:\qquad {\tfrac {1}{2}}\left(5({\tfrac {29}{5}})-5\right)=0,\quad {\tfrac {1}{2}}\left(5({\tfrac {29}{5}})+5\right)=5.}$

${\displaystyle F_{14}=377{\text{ and }}F_{15}=610.}$

${\displaystyle 5F_{14}^{2}=710645\equiv 0{\pmod {29}}\;\;{\text{ and }}\;\;5F_{15}^{2}=1860500\equiv 5{\pmod {29}}}$

For odd n, all odd prime divisors of F_n are congruent to 1 modulo 4, implying that all odd divisors of F_n (as the products of odd prime divisors) are congruent to 1 modulo 4.[82]

For example,

${\displaystyle F_{1}=1,F_{3}=2,F_{5}=5,F_{7}=13,F_{9}=34=2\cdot 17,F_{11}=89,F_{13}=233,F_{15}=610=2\cdot 5\cdot 61.}$

All known factors of Fibonacci numbers F(i) for all i < 50000 are collected at the relevant repositories.[83][84]

If the members of the Fibonacci sequence are taken mod n, the resulting sequence is periodic with period at most 6n.[85] The lengths of the periods for various n form the so-called Pisano periods OEIS: A001175. 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.

Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple. The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle.

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

${\displaystyle {\begin{aligned}a_{n}&=F_{2n-1}\\[4pt]b_{n}&=2F_{n}F_{n-1}\\[4pt]c_{n}&=F_{n}^{2}-F_{n-1}^{2}=F_{n+1}F_{n-2}.\end{aligned}}}$

These formulas satisfy ${\displaystyle a_{n}^{2}=b_{n}^{2}+c_{n}^{2}}$ for all n, but they only represent triangle sides when n > 2.

Any four consecutive Fibonacci numbers F_n, F_n+1, F_n+2 and F_n+3 can also be used to generate a Pythagorean triple in a different way:[86]

${\displaystyle {\begin{aligned}a&=F_{n}F_{n+3}\\b&=2F_{n+1}F_{n+2}\\c&=F_{n+1}^{2}+F_{n+2}^{2}.\end{aligned}}}$

Since F_n is asymptotic to ${\displaystyle \varphi ^{n}/{\sqrt {5}}}$, the number of digits in F_n is asymptotic to ${\displaystyle n\log _{10}\varphi \approx 0.2090\,n}$. 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 F_n is asymptotic to ${\displaystyle n\log _{b}\varphi .}$

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, from Fibonacci sequence include:

• Generalizing the index to negative integers to produce the negafibonacci numbers.
• Generalizing the index to real numbers using a modification of Binet's formula.[60]
• Starting with other integers. Lucas numbers have L₁ = 1, L₂ = 3, and L_n = L_n−1 + L_n−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. The Pell numbers have P_n = 2P_{n − 1} + P_{n − 2}. If the coefficient of the preceding value is assigned a variable value x, the result is the sequence of Fibonacci polynomials.
• Not adding the immediately preceding numbers. The Padovan sequence and Perrin numbers have P(n) = P(n − 2) + P(n − 3).
• Generating the next number by adding 3 numbers (tribonacci numbers), 4 numbers (tetranacci numbers), or more. The resulting sequences are known as n-Step Fibonacci numbers.[87]