Fibonacci number

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A tiling with squares whose sides are successive Fibonacci numbers in length

A Fibonacci spiral created by drawing circular arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34. See golden spiral.

In mathematics, the Fibonacci numbers are the numbers in the following integer sequence:

$0,\;1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;144,\; \ldots\;$ (sequence A000045 in OEIS).

By definition, the first two numbers in the Fibonacci sequence are 0 and 1, and each subsequent number is the sum of the previous two.

In mathematical terms, the sequence F_n of Fibonacci numbers is defined by the recurrence relation

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

with seed values^[1]

$F_0 = 0,\; F_1 = 1$

The Fibonacci sequence is named after Leonardo of Pisa, who was known as Fibonacci. Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics,^[2] although the sequence had been described earlier in Indian mathematics.^[3]^[4]^[5] (By modern convention, the sequence begins with F₀ = 0. The Liber Abaci began the sequence with F₁ = 1, omitting the initial 0, and the sequence is still written this way by some.)

Fibonacci numbers are closely related to Lucas numbers in that they are a complementary pair of Lucas sequences. They are intimately connected with the golden ratio, for example the closest rational approximations to the ratio are 2/1, 3/2, 5/3, 8/5, ... . Applications 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,^[6] such as branching in trees, arrangement of leaves on a stem, the fruit spouts of a pineapple,^[7] the flowering of artichoke, an uncurling fern and the arrangement of a pine cone.^[8]

1 Origins
2 List of Fibonacci numbers
3 Occurrences in mathematics
4 Relation to the golden ratio
- 4.1 Closed-form expression
- 4.2 Computation by rounding
- 4.3 Limit of consecutive quotients
- 4.4 Decomposition of powers of the golden ratio
5 Matrix form
6 Recognizing Fibonacci numbers
7 Identities
- 7.1 First identity
- 7.2 Second identity
- 7.3 Third identity
- 7.4 Fourth identity
- 7.5 Fifth identity
- 7.6 Identity for doubling n
- 7.7 Another identity
- 7.8 Other identities
8 Power series
9 Reciprocal sums
10 Primes and divisibility
- 10.1 Divisibility properties
- 10.2 Fibonacci primes
- 10.3 Prime divisors of Fibonacci numbers
- 10.4 Periodicity modulo n
11 Right triangles
12 Magnitude
13 Applications
14 In nature
- 14.1 The bee ancestry code
15 Popular culture
16 Generalizations
17 See also
18 Notes
19 References
20 External links

[ Origins

The Fibonacci sequence appears in Indian mathematics, in connection with Sanskrit prosody.^[4]^[9] In the Sanskrit oral tradition, there was much emphasis on how long (L) syllables mix with the short (S), and counting the different patterns of L and S within a given fixed length results in the Fibonacci numbers; the number of patterns that are m short syllables long is the Fibonacci number F_m + 1.^[5]

Susantha Goonatilake writes that the development of the Fibonacci sequence "is attributed in part to Pingala (200 BC), later being associated with Virahanka (c. 700 AD), Gopāla (c.1135 AD), and Hemachandra (c.1150)".^[3] Parmanand Singh cites Pingala's cryptic formula misrau cha ("the two are mixed") and cites scholars who interpret it in context as saying that the cases for m beats (F_m+1) is obtained by adding a [S] to F_m cases and [L] to the F_m−1 cases. He dates Pingala before 450 BCE.^[10]

However, the clearest exposition of the series arises in the work of Virahanka (c. 700AD), whose own work is lost, but is available in a quotation by Gopala (c.1135):

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 mAtrA-vr.ttas (prosodic combinations).^[11]

The series is also discussed by Gopala (before 1135AD) and by the Jain scholar Hemachandra (c. 1150AD).

In the West, the Fibonacci sequence first appears in the book Liber Abaci (1202) by Leonardo of Pisa, known as Fibonacci.^[2] Fibonacci considers the growth of an idealized (biologically unrealistic) rabbit population, assuming that: a newly born pair of rabbits, one male, one female, are put in a field; rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits; rabbits never die and a mating pair always produces one new pair (one male, one female) every month from the second month on. The puzzle that Fibonacci posed was: 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 the female produces a new pair, so now there are 2 pairs of rabbits in the field.
At the end of the third month, the original female produces a second pair, making 3 pairs in all in the field.
At the end of the fourth month, the original female has produced yet another new pair, the female born two months ago produces her first pair also, making 5 pairs.

At the end of the nth month, the number of pairs of rabbits is equal to the number of new pairs (which is the number of pairs in month n − 2) plus the number of pairs alive last month (n − 1). This is the nth Fibonacci number.^[12]

The name "Fibonacci sequence" was first used by the 19th-century number theorist Édouard Lucas.^[13]

[ List of Fibonacci numbers

The first 21 Fibonacci numbers F_n for n = 0, 1, 2, ..., 20 are:[14]

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

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

which yields the sequence of "negafibonacci" numbers^[15] satisfying

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

Thus the complete 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

[ Occurrences in mathematics

The Fibonacci numbers are the sums of the "shallow" diagonals (shown in red) of Pascal's triangle.

The Fibonacci numbers occur in the sums of "shallow" diagonals in Pascal's triangle (see Binomial coefficient).^[16]

The Fibonacci numbers can be found in different ways in the sequence of binary strings.

The number of binary strings of length n without consecutive 1s is the Fibonacci number F_n+2. For example, out of the 16 binary strings of length 4, there are F₆ = 8 without consecutive 1s – they are 0000, 0100, 0010, 0001, 0101, 1000, 1010 and 1001. By symmetry, the number of strings of length n without consecutive 0s is also F_n+2.
The number of binary strings of length n without an odd number of consecutive 1s is the Fibonacci number F_n+1. For example, out of the 16 binary strings of length 4, there are F₅ = 5 without an odd number of consecutive 1s – they are 0000, 0011, 0110, 1100, 1111.
The number of binary strings of length n without an even number of consecutive 0s or 1s is 2F_n. For example, out of the 16 binary strings of length 4, there are 2F₄ = 6 without an even number of consecutive 0s or 1s – they are 0001, 1000, 1110, 0111, 0101, 1010.

[ Relation to the golden ratio

[ Closed-form expression

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, even though it was already known by Abraham de Moivre:^[17]

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

where

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

is the golden ratio (sequence A001622 in OEIS), and

$\psi = \frac{1 - \sqrt{5}}{2} = 1 - \varphi = - {1 \over \varphi}.$ ^[18]

To see this,^[19] note that φ and ψ are both solutions of the equations

$x^2 = x + 1,\, x^n = x^{n-1} + x^{n-2},\,$

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

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

and

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

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

$U_n=a \varphi^n + b \psi^n\,$

satisfies the same recurrence

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

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:

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

which has solution

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

producing the required formula.

[ Computation by rounding

Since

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

for all n ≥ 0, the number F_n is the closest integer to

$\frac{\varphi^n}{\sqrt 5}\, .$

Therefore it can be found by rounding, or in terms of the floor function:

$F_n=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor,\ n \geq 0.$

Similarly, if we already know that the number F > 1 is a Fibonacci number, we can determine its index within the sequence by

$n(F) = \bigg\lfloor \log_\varphi \left(F\cdot\sqrt{5} + \frac{1}{2}\right) \bigg\rfloor$

[ 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 the limit approaches the golden ratio $\varphi\,$ .^[20]

$\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi$

This convergence does not depend on the starting values chosen, excluding 0, 0. For example, the initial values 19 and 31 generate the sequence 19, 31, 50, 81, 131, 212, 343, 555 ... etc. The ratio of consecutive terms in this sequence shows the same convergence towards the golden ratio.

In fact this holds for any sequence which satisfies the Fibonacci recurrence other than a sequence of 0's. This can be derived from Binet's formula.

[ Decomposition of powers of the golden ratio

Since the golden ratio satisfies the equation

$\varphi^2 = \varphi + 1,\,$

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

φ n = F (n)φ + F (n - 1).

This expression is also true for $n \, <\, 1 \,$ if the Fibonacci sequence $F(n) \,$ is extended to negative integers using the Fibonacci rule $F(n) = F(n-1) + F(n-2) . \,$

[ Matrix form

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

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

The eigenvalues of the matrix A are $\scriptstyle \varphi\,\!$ and $\scriptstyle (1-\varphi)\,\!$ , and the elements of the eigenvectors of A, $\scriptstyle {\varphi \choose 1}$ and $\scriptstyle {1 \choose -\varphi}$ , are in the ratios $\scriptstyle \varphi\,\!$ and $(1-\varphi\,\!).$ Using these facts, and the properties of eigenvalues, we can derive a direct formula for the nth element in the Fibonacci series:

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

The matrix has a determinant of −1, and thus it is a 2×2 unimodular matrix. This property can be understood in terms of the continued fraction representation for the golden ratio:

$\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\;\;\ddots\,}}}$

The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for $\varphi\,\!$ , and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1.

The matrix representation gives the following closed expression for the Fibonacci numbers:

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

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

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

Additionally, since $A n A m = A m + n$ for any square matrix A, the following identities can be derived:

$\begin{align} {F_m}{F_n} + {F_{m-1}}{F_{n-1}} &= F_{m+n-1}\\ F_{n+1}F_{m} + F_n F_{m-1} &= F_{m+n} \end{align}$

In particular, with $m = n$ ,

$\begin{align} F_{2n-1} &= F_n^2 + F_{n-1}^2\\ F_{2n} &= (F_{n-1}+F_{n+1})F_n\\ &= (2F_{n-1}+F_n)F_n \end{align}$

[ Recognizing Fibonacci numbers

The question may arise whether a positive integer z is a Fibonacci number. Since $F (n)$ is the closest integer to $\varphi^n/\sqrt{5}$ , the most straightforward, brute-force test is the identity

$F\bigg(\bigg\lfloor\log_\varphi\bigg(z\cdot\sqrt{5}+\frac{1}{2}\bigg)\bigg\rfloor\bigg)=z,$

which is true if and only if z is a Fibonacci number. In this formula, $F (n)$ can be computed rapidly using any of the previously discussed closed-form expressions.

One implication of the above expression is this: if it is known that a number z is a Fibonacci number, we may determine an n such that F(n) = z by the following:

$\bigg\lfloor\log_\varphi\bigg(z\cdot\sqrt{5}+\frac{1}{2}\bigg)\bigg\rfloor = n$

Alternatively, a positive integer z is a Fibonacci number if and only if one of $5 z 2 + 4$ or $5 z 2 - 4$ is a perfect square.^[21]

A slightly more sophisticated test uses the fact that the convergents of the continued fraction representation of $\varphi\,$ are ratios of successive Fibonacci numbers. That is, the inequality

$\bigg|\varphi-\frac{p}{q}\bigg|<\frac{1}{q^2}$

(with coprime positive integers p, q) is true if and only if p and q are successive Fibonacci numbers. From this one derives the criterion that z is a Fibonacci number if and only if the closed interval

$\bigg[\varphi z-\frac{1}{z},\varphi z+\frac{1}{z}\bigg]$

contains a positive integer.^[22] For $z \geq 2$ , it is easy to show that this interval contains at most one integer, and in the event that z is a Fibonacci number, the contained integer is equal to the next successive Fibonacci number after z. Somewhat remarkably, this result still holds for the case $z = 1$ , but it must be stated carefully since $1$ appears twice in the Fibonacci sequence, and thus has two distinct successors.

[ Identities

Most identities involving Fibonacci numbers draw from combinatorial arguments. F(n) can be interpreted as the number of sequences of 1s and 2s that sum to n − 1, with the convention that F(0) = 0, meaning no sum will add up to −1, and that F(1) = 1, meaning the empty sum will "add up" to 0. Here the order of the summands matters. For example, 1 + 2 and 2 + 1 are considered two different sums and are counted twice.

[ First identity

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

For n > 1.

The nth Fibonacci number is the sum of the previous two Fibonacci numbers.

Proof

We must establish that the sequence of numbers defined by the combinatorial interpretation above satisfy the same recurrence relation as the Fibonacci numbers (and so are indeed identical to the Fibonacci numbers).

The set of F(n + 1) ways of making ordered sums of 1s and 2s that sum to n may be divided into two non-overlapping sets. The first set contains those sums whose first summand is 1; the remainder sums to n − 1, so there are F(n) sums in the first set. The second set contains those sums whose first summand is 2; the remainder sums to n − 2, so there are F(n − 1) sums in the second set. The first summand can only be 1 or 2, so these two sets exhaust the original set. Thus F(n + 1) = F(n) + F(n − 1).

[ Second identity

The sum of the first n Fibonacci numbers is equal to the n+2nd Fibonacci number minus 1.[23] In symbols:

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

The sum of the first n Fibonacci numbers is the (n + 2)nd Fibonacci number minus 1.

Proof

We count the number of ways summing 1s and 2s to n + 1 such that at least one of the summands is 2.

As before, there are F(n + 2) ways summing 1s and 2s to n + 1 when n ≥ 0. Since there is only one sum of n + 1 that does not use any 2, namely 1 + ... + 1 (n + 1 terms), we subtract 1 from F(n + 2).

Equivalently, we can consider the first occurrence of 2 as a summand. If, in a sum, the first summand is 2, then there are F(n) ways to the complete the counting for n − 1. If the second summand is 2 but the first is 1, then there are F(n − 1) ways to complete the counting for n − 2. Proceed in this fashion. Eventually we consider the (n + 1)th summand. If it is 2 but all of the previous n summands are 1s, then there are F(0) ways to complete the counting for 0. If a sum contains 2 as a summand, the first occurrence of such summand must take place in between the first and (n + 1)th position. Thus F(n) + F(n − 1) + ... + F(0) gives the desired counting.

By induction:

For

n = 0

, $\sum_{i=0}^0 F_i = F_2 - 1 = 1 - 1 = 0$ , so the equation is true for

n = 0

For

n = x

, assume $\sum_{i=0}^x F_i = F_{x+2} - 1$ .

Add the next Fibonacci number

F x + 1

to both sides: $F_{x+1} + \sum_{i=0}^x F_i = F_{x+1} + F_{x+2} - 1$ .

By the Fibonacci recurrence relation,

F x + 1 + F x + 2 = F x + 3

, so $\sum_{i=0}^{x+1} F_i = F_{x+3} - 1$ , which is the

n = x + 1

case, proving that where the equation is true for

n = x

, so is it for

n = x + 1

[ Third identity

This identity has slightly different forms for F_j, depending on whether j is odd or even.

The sum of the first n − 1 Fibonacci numbers, F_j, such that j is odd, is the (2n)th Fibonacci number.

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

The sum of the first n Fibonacci numbers, F_j, such that j is even, is the (2n + 1)th Fibonacci number minus 1.

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

[24]

Proofs

1: j is odd

By induction for F_2n:

$F_1+F_3+F_5+\cdots+F_{2n-3}+F_{2n-1}=F_{2n}\,$

$F_1+F_3+F_5+\cdots+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n}+F_{2n+1}\,$

$F_1+F_3+F_5+\cdots+F_{2n-3}+F_{2n-1}+F_{2n+1}=F_{2n+2}\,$

A basis case for this could be F₁ = F₂.

2: j is even

By induction for F_2n+1:

$F_0+F_2+F_4+\cdots+F_{2n-2}+F_{2n}=F_{2n+1}-1\,$

$F_0+F_2+F_4+\cdots+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+1}+F_{2n+2}-1\,$

$F_0+F_2+F_4+\cdots+F_{2n-2}+F_{2n}+F_{2n+2}=F_{2n+3}-1\,$

A basis case for this could be F₀ = F₁ − 1.

Alternative proof

By using identity 1 we can construct a telescoping sum:

$\sum_{i=0}^{n-1} F_{2i+1} = \sum_{i=0}^{n-1} [ F_{2(i+1)}-F_{2i} ] = F_{2n}-F_{0} = F_{2n}$

If the summands are the Fibonacci numbers with even index, the proof is very similar. Summing both cases yields identity 2.

[ Fourth identity

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

Proof

This identity can be established in two stages. First, we count the number of ways summing 1s and 2s to −1, 0, ..., or n + 1 such that at least one of the summands is 2.

By our second identity, there are F(n + 2) − 1 ways summing to n + 1; F(n + 1) − 1 ways summing to n; ...; and, eventually, F(2) − 1 way summing to 1. As F(1) − 1 = F(0) = 0, we can add up all n + 1 sums and apply the second identity again to obtain

[F(n + 2) − 1] + [F(n + 1) − 1] + ... + [F(2) − 1]

= [F(n + 2) − 1] + [F(n + 1) − 1] + ... + [F(2) − 1] + [F(1) − 1] + F(0)

= F(n + 2) + [F(n + 1) + ... + F(1) + F(0)] − (n + 2)

= F(n + 2) + [F(n + 3) − 1] − (n + 2)

= F(n + 2) + F(n + 3) − (n + 3).

On the other hand, we observe from the second identity that there are

F(0) + F(1) + ... + F(n − 1) + F(n) ways summing to n + 1;
F(0) + F(1) + ... + F(n − 1) ways summing to n;

......

F(0) way summing to −1.

Adding up all n + 1 sums, we see that there are

(n + 1) F(0) + n F(1) + ... + F(n) ways summing to −1, 0, ..., or n + 1.

Since the two methods of counting refer to the same number, we have

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

Finally, we complete the proof by subtracting the above identity from n + 1 times the second identity.

[ Fifth identity

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

The sum of the squares of the first n Fibonacci numbers is the product of the nth and (n + 1)th Fibonacci numbers.

Proof

Although this identity can be established by either induction or direct, albeit messy, algebraic manipulation, perhaps the most elegant and most insightful method is by a simple geometric argument.

Consider the Fibonacci Rectangles constructed in previous sections. Using a common trick, we will compute the area of this rectangle in two different ways. But since this must yield the same answer in both cases, we know these resulting expressions must be equal, which will yield the desired identity.

On the one hand, the n-th rectangle is composed of n squares, whose side lengths are F(1), F(2), ..., F(n). Its area is therefore the sum of each of these squares, which is given by

$\sum_{i=0}^n {F_i}^2.$

On the other hand, we know that the n-th rectangle has side lengths F(n) and F(n + 1). Thus, its area is simply given by

$F_{n} F_{n+1}. \,$

Setting these expressions equal to each other completes the proof.

[ Identity for doubling n

$F_{2n} = F_{n+1}^2 - F_{n-1}^2 = F_n(F_{n+1}+F_{n-1}) = F_nL_n$ [25]

Where $L n$ is the n'th Lucas Number.

[ Another identity

Another identity useful for calculating F_n for large values of n is [25]

$F_{kn+c} = \sum_{i=0}^k {k\choose i} F_{c-i} F_n^i F_{n+1}^{k-i},$

from which other identities for specific values of k, n, and c can be derived below, including

$F_{2n+k} = F_k F_{n+1}^2 + 2 F_{k-1} F_{n+1} F_n + F_{k-2} F_n^2$

for all integers n and k. Doubling identities of this type can be used to calculate F_n using O(log n) long multiplication operations of size n bits. The number of bits of precision needed to perform each multiplication doubles at each step, so the performance is limited by the final multiplication; if the fast Schönhage–Strassen multiplication algorithm is used, this is O(n log n log log n) bit operations. Notice that, with the definition of Fibonacci numbers with negative n given in the introduction, this formula reduces to the double n formula when k = 0.

[ Other identities

Other identities include relationships to the Lucas numbers, which have the same recursive properties but start with L₀ = 2 and L₁ = 1. These properties include F_2n = F_nL_n.

There are also scaling identities, which take you from F_n and F_n+1 to a variety of things of the form F_an+b; for instance

$F_{3n} = 2F_n^3 + 3F_n F_{n+1} F_{n-1} = 5F_n^3 + 3 (-1)^n F_n \,$ by Cassini's identity.

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

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

$F_{4n} = 4F_nF_{n+1}(F_{n+1}^2 + 2F_n^2) - 3F_n^2(F_n^2 + 2F_{n+1}^2) \,$

These can be found experimentally using lattice reduction, and are useful in setting up the special number field sieve to factorize a Fibonacci number. Such relations exist in a very general sense for numbers defined by recurrence relations. See the section on multiplication formulae under Perrin numbers for details.

[ Power series

The generating function of the Fibonacci sequence is the power series

$s(x)=\sum_{k=0}^{\infty} F_k x^k.$

This series has a simple and interesting closed-form solution for $|x| < \frac{1}{\varphi}$ :^[26]

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

This solution can be proven by using the Fibonacci recurrence to expand each coefficient in the infinite sum defining $s (x)$ :

$\begin{align} s(x) &= \sum_{k=0}^{\infty} F_k x^k \\ &= F_0 + F_1x + \sum_{k=2}^{\infty} \left( F_{k-1} + F_{k-2} \right) x^k \\ &= x + \sum_{k=2}^{\infty} F_{k-1} x^k + \sum_{k=2}^{\infty} F_{k-2} x^k \\ &= x + x\sum_{k=0}^{\infty} F_k x^k + x^2\sum_{k=0}^{\infty} F_k x^k \\ &= x + x s(x) + x^2 s(x). \end{align}$

Solving the equation $s (x) = x + x s (x) + x 2 s (x)$ for $s (x)$ results in the closed form solution.

In particular, math puzzle-books note the curious value $\frac{s(\frac{1}{10})}{10}=\frac{1}{89}$ ,^[27] or more generally

$\sum_{n = 1}^{\infty}{\frac {F_n}{10^{(k + 1)(n + 1)}}} = \frac {1}{10^{2k + 2} - 10^{k + 1} - 1}$

for all integers $k \geq 0$ .

More generally,

$\sum_{n=0}^\infty\,\frac{F_n}{k^{n}}\,=\,\frac{k}{k^{2}-k-1}.$

[ Reciprocal sums

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

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

and the sum of squared reciprocal Fibonacci numbers as

$\sum_{k=1}^\infty \frac{1}{F_k^2} = \frac{5}{24} \left(\vartheta_2^4\left(0, \frac{3-\sqrt 5}{2}\right) - \vartheta_4^4\left(0, \frac{3-\sqrt 5}{2}\right) + 1 \right).$

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

$\sum_{k=0}^\infty \frac{1}{1+F_{2k+1}} = \frac{\sqrt{5}}{2},$ CC-BY-SA.

Fibonacci sequence

Fibonacci number

Contents

[ Origins

[ List of Fibonacci numbers

[ Occurrences in mathematics

[ Relation to the golden ratio

[ Closed-form expression

[ Computation by rounding

[ Limit of consecutive quotients

[ Decomposition of powers of the golden ratio

[ Matrix form

[ Recognizing Fibonacci numbers

[ Identities

[ First identity

[ Second identity

[ Third identity

[ Fourth identity

[ Fifth identity

[ Identity for doubling n

[ Another identity

[ Other identities

[ Power series

[ Reciprocal sums

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