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Fibonacci Sequence



Definition
Fibonacci Notation
The Golden Ratio
Binet's Formula
Identities and Properties
Lucas Numbers
Fibonacci at a Glance
Common Pitfalls
Fibonacci Sequence FAQ



Each Term from the Two Before It

The Fibonacci sequence is the most famous example of a recursively defined sequence: each term is the sum of the two terms that precede it. From the simplest possible starting values — 11 and 11 — this rule produces a sequence whose growth rate, divisibility structure, and algebraic identities have been studied for centuries. Despite its purely recursive origin, the Fibonacci sequence has an explicit closed-form expression involving the golden ratio.



Definition

A pair of young rabbits grows up in a month and from then on produces one new pair every month. How many pairs are there month by month?
This puzzle appears in Leonardo of Pisa's book of 1202, the Liber Abaci, and Leonardo is better known today as Fibonacci. No rabbit ever dies in the puzzle; every new pair follows the same rule.
Counting month by month gives 1, 1, 2, 3, 5, 8 pairs.
monthadult pairsyoung pairstotal pairs
1011
2101
3112
4213
5325
6538
Every adult pair of one month is still there the next, and every pair that was already adult a month earlier has just added a young pair.
Each month's total is the sum of the two totals before it.
The pairs alive last month are all still alive, and the pairs alive two months ago are now old enough to have one new pair each: 8=5+38 = 5 + 3.
Each term after the second is the sum of its two immediate predecessors: 2=1+12 = 1 + 1, 3=1+23 = 1 + 2, 5=2+35 = 2 + 3, 8=3+58 = 3 + 5, and so on. Growth accelerates because each new term absorbs the size of both parents.
The definition
The Fibonacci sequence starts with 1,11, 1 and continues by adding the two previous terms: Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2}.
The Fibonacci sequence is defined by a two-term recurrence with two initial values:
Fibonacci Recurrence
F1=1,F2=1,Fn=Fn−1+Fn−2(n≥3)F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \quad (n \geq 3)
Learn more about this formula: Fibonacci Recurrence →
The first terms are:
Definition
1,1,2,3,5,8,13,21,34,55,89,144,233,377,…1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, \ldots
So the rabbit counts 1,1,2,3,5,81, 1, 2, 3, 5, 8 are F1F_1 to F6F_6.
The Fibonacci sequence is defined recursively, and it is neither arithmetic nor geometric.
The Fibonacci sequence is neither arithmetic (the differences 0,1,1,2,3,5,…0, 1, 1, 2, 3, 5, \ldots are not constant) nor geometric (the ratios 1,2,1.5,1.6‾,1.6,…1, 2, 1.5, 1.\overline{6}, 1.6, \ldots are not constant either, though they converge).
To compute any Fibonacci number or list a range of terms, use the Fibonacci sequence calculator.
Two starting conventions are in use, and they shift every index by one.
An alternative convention starts with F0=0F_0 = 0: the sequence becomes 0,1,1,2,3,5,8,…0, 1, 1, 2, 3, 5, 8, \ldots. The same recurrence applies and every FnF_n keeps its value — only the list gains a 00 in front, so each term sits one place later in the list. Both conventions appear in the literature, and the choice is a matter of context. This page uses F1=1F_1 = 1.
Pitfall: Reading the list position as the index
Taking the 10th entry of 0,1,1,2,3,5,8,13,21,34,…0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots as F10F_{10} gives F10=34F_{10} = 34. That list starts at F0F_0, so its 10th entry is F9=34F_9 = 34, and F10=55F_{10} = 55.
The two lists differ only by the extra 00 in front, and that difference matters as soon as a term is called by its number. Binet's formula and the identities in the sections below are all stated in terms of the index nn of FnF_n, with F1=F2=1F_1 = F_2 = 1.
The rest of the page covers the golden ratio, Binet's formula, identities and properties, and the Lucas numbers.

Fibonacci Notation

Notation

Fibonacci Notation

A capital letter kept for one sequence, two Greek letters for the roots behind it, and a second capital for its twin.
The subscript as position, the brace form and sigma are set out at sequence notation; gcd⁡\gcd in the identities is read as in GCD notation.
FnF_n
F sub n — the n-th Fibonacci number
Capital FF is reserved for this one sequence, the way ana_n serves any sequence. This page counts from F1=F2=1F_1 = F_2 = 1, and every identity below uses that index.
CasesThe other convention starts at F0=0F_0 = 0. Values do not change — F10=55F_{10} = 55 either way — but the list gains a leading 00, so its 10th entry is F9F_9.
Also writtenF(n)F(n), fib(n)\text{fib}(n) — function and code forms.
Do not confuseList position. In a list that starts at F0F_0, the k-th entry is Fk−1F_{k-1} — read the index, not the place in line.
ϕ,  ψ\phi, \; \psi
phi and psi — the golden ratio and its conjugate
The two roots of x2=x+1x^2 = x + 1: ϕ=1+52≈1.618\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618 and ψ=1−52≈−0.618\psi = \frac{1 - \sqrt{5}}{2} \approx -0.618. Binet's formula writes FnF_n from their powers.
Casesψ\psi is negative — dropping its sign turns Binet's formula into non-integers. The pair satisfies ϕ+ψ=1\phi + \psi = 1 and ϕψ=−1\phi\psi = -1.
Also writtenφ\varphi — the curly form of phi, same number. ϕˉ\bar{\phi} or ϕ′\phi' for the second root in some books.
Do not confuseThe smaller golden number 1ϕ=ϕ−1≈0.618\frac{1}{\phi} = \phi - 1 \approx 0.618, which some texts also call the golden ratio. Note ψ=−1ϕ\psi = -\frac{1}{\phi}: same digits, opposite sign.
LnL_n
L sub n — the n-th Lucas number
The same recurrence Ln=Ln−1+Ln−2L_n = L_{n-1} + L_{n-2} with a different start, L1=1L_1 = 1 and L2=3L_2 = 3: 1,3,4,7,11,…1, 3, 4, 7, 11, \ldots (Lucas Numbers). In Binet form the minus becomes a plus: Ln=ϕn+ψnL_n = \phi^n + \psi^n.
CasesTied to the Fibonacci numbers by Ln=Fn−1+Fn+1L_n = F_{n-1} + F_{n+1}: L5=F4+F6=11L_5 = F_4 + F_6 = 11.
Do not confuseFnF_n itself — the two letters share every step of the rule and differ only in the second starting value.
DownloadPNG2×PDF
Notation · Cheat sheet
Sequence Notation at a Glance
The marks from the notation cards across the sequence pages: how each is read and what it means. Every group links to the full card on its page.
Sequences
ana_n
a sub n
The term in place nn; the subscript is a position.
a1,a2,a3,…a_1, a_2, a_3, \ldots
a one, a two, a three, and so on
The terms listed in order; the dots need a stated rule.
{an}n=1∞\{a_n\}_{n=1}^{\infty}
the sequence a sub n
The whole sequence as one object.
Arithmetic sequences
an+1=an+da_{n+1} = a_n + d
next term is this term plus d
The recurrence; dd is the common difference, a1a_1 the start.
∑k=1nak\textstyle\sum_{k=1}^{n} a_k
sum of a sub k, k from 1 to n
Sigma: the sum SnS_n of the first nn terms.
Geometric sequences
rr
r, the common ratio
The factor from each term to the next: an=a1rn−1a_n = a_1 r^{n-1}.
SnS_n
S sub n
Sum of the first nn terms; nn counts terms.
SS
S, the infinite sum
Equals a1/(1−r)a_1 / (1 - r), only when ∣r∣<1|r| < 1.
GG
G, the geometric mean
G=abG = \sqrt{ab}, the middle value of a,G,ba, G, b.
Fibonacci sequenceThis page
FnF_n
F sub n
The nn-th Fibonacci number, counted from F1=F2=1F_1 = F_2 = 1.
ϕ,  ψ\phi, \; \psi
phi and psi
The roots of x2=x+1x^2 = x + 1, about 1.6181.618 and −0.618-0.618.
LnL_n
L sub n
Lucas numbers: the same rule, started at 1,31, 3.
Harmonic sequences
an=1/bna_n = 1/b_n
a sub n is one over b sub n
bnb_n is the arithmetic sequence underneath.
HnH_n
H sub n
The harmonic number 1+12+⋯+1n1 + \frac{1}{2} + \cdots + \frac{1}{n}.
HH
H, the harmonic mean
For two numbers, H=2aba+bH = \frac{2ab}{a + b}.
Triangular numbers
TnT_n
T sub n
The nn-th triangular number, n(n+1)2\frac{n(n+1)}{2}.
Tn=Tn−1+nT_n = T_{n-1} + n
the one before plus a row of n
The recurrence, from T1=1T_1 = 1.
(n+12)\binom{n+1}{2}
n plus one choose two
The same number, counted as pairs.
Square numbers
Sn=n2S_n = n^2
S sub n on this page
The nn-th square: one term, not a sum.
Sn=Sn−1+2n−1S_n = S_{n-1} + 2n - 1
the one before plus the next odd number
The recurrence, from S1=1S_1 = 1.
∑k=1nk2\textstyle\sum_{k=1}^{n} k^2
sum of k squared
Square each term first, then add.
Prime numbers
pp
p, a prime
The letter promises a prime.
p1,…,pkp_1, \ldots, p_k
p sub one to p sub k
A list of primes, or the primes of one number.
π(n)\pi(n)
pi of n
How many primes are at most nn.
Same letter, different job
SnS_nA sum of nn terms on the arithmetic and geometric pages; a single square on the square numbers page.
HnH_n vs HHWith a subscript, a sum of reciprocals; without one, a mean.
π(n)\pi(n) vs π\piA count of primes, not the circle constant.
{an}\{a_n\}An ordered sequence, not a set: terms keep their order and may repeat.
Sequence symbols tableSequences
Sequence Notation at a Glance·/algebra/sequencesLearn Math Class

The Golden Ratio

The ratio of consecutive Fibonacci numbers Fn+1Fn\frac{F_{n+1}}{F_n} settles toward a fixed value as nn grows:

The Golden Ratio
F2F1=1,F3F2=2,F4F3=1.5,F5F4=1.6‾,F6F5=1.6,…\frac{F_2}{F_1} = 1, \quad \frac{F_3}{F_2} = 2, \quad \frac{F_4}{F_3} = 1.5, \quad \frac{F_5}{F_4} = 1.\overline{6}, \quad \frac{F_6}{F_5} = 1.6, \quad \ldots

The limit is the golden ratio:

Golden Ratio
ϕ=1+52≈1.6180339887…\phi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887\ldots
Learn more about this formula: Golden Ratio →


The number ϕ\phi satisfies the equation ϕ2=ϕ+1\phi^2 = \phi + 1, or equivalently ϕ2−ϕ−1=0\phi^2 - \phi - 1 = 0. This quadratic has a second root ψ=1−52≈−0.618\psi = \frac{1 - \sqrt{5}}{2} \approx -0.618, which also satisfies ψ2=ψ+1\psi^2 = \psi + 1.

Pitfall: Dividing the earlier term by the later one
Taking FnFn+1\frac{F_n}{F_{n+1}} instead gives 5589≈0.618\frac{55}{89} \approx 0.618, which is 1ϕ=ϕ−1\frac{1}{\phi} = \phi - 1, not ϕ\phi. The ratio is later over earlier: 8955≈1.618\frac{89}{55} \approx 1.618.

The two roots are related: ϕ+ψ=1\phi + \psi = 1 and ϕ⋅ψ=−1\phi \cdot \psi = -1. Since ∣ψ∣<1|\psi| < 1, powers of ψ\psi shrink toward zero, which is why the ratio Fn+1Fn\frac{F_{n+1}}{F_n} converges to ϕ\phi and not to ψ\psi.
Ratios of consecutive Fibonacci numbers 1, 2, 1.5, 1.667, 1.6, ... close in on phi from alternating sidesny123456789101112φ ≈ 1.6182/11/13/25/38/5Fₙ₊₁ / Fₙ alternates above and below φ, closing inThe ratios of consecutive Fibonacci numbers converge to φ
Ratios settling on the golden ratio

Each dot is the ratio Fₙ₊₁/Fₙ: 1/1, 2/1, 3/2, 5/3, 8/5, and so on. The ratios jump above and below the dashed level φ ≈ 1.618, and each jump is smaller than the last. By n = 6 the dots sit on the line to the eye; 89/55 = 1.61818… already agrees with φ to three decimal places.

The alternation comes from ψ: its powers flip sign while shrinking toward 0.

Binet's Formula

Despite its recursive definition, the Fibonacci sequence has an explicit closed-form expression:

Binet's Formula
Fn=ϕn−ψn5F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}
Learn more about this formula: Binet's Formula →


where ϕ=1+52\phi = \frac{1 + \sqrt{5}}{2} and ψ=1−52\psi = \frac{1 - \sqrt{5}}{2}.

The derivation starts by assuming a solution of the form Fn=xnF_n = x^n and substituting into the recurrence Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2}. This gives xn=xn−1+xn−2x^n = x^{n-1} + x^{n-2}, which simplifies to the characteristic equation x2=x+1x^2 = x + 1. The two roots are ϕ\phi and ψ\psi. Since the recurrence is linear, any linear combination Fn=Aϕn+BψnF_n = A\phi^n + B\psi^n is also a solution, and the constants AA and BB are determined by the initial conditions F1=1F_1 = 1, F2=1F_2 = 1.

Pitfall: Dropping the sign of the second root
Using ψ≈+0.618\psi \approx +0.618 gives ψ3≈0.236\psi^3 \approx 0.236 and, with ϕ3≈4.236\phi^3 \approx 4.236, F3=4.236−0.2365≈1.789F_3 = \frac{4.236 - 0.236}{\sqrt{5}} \approx 1.789, not a whole number. With ψ≈−0.618\psi \approx -0.618, ψ3≈−0.236\psi^3 \approx -0.236 and F3=4.236+0.2365=2F_3 = \frac{4.236 + 0.236}{\sqrt{5}} = 2.

The formula involves irrational numbers — 5\sqrt{5} appears explicitly, and both ϕ\phi and ψ\psi are irrational — yet it always produces an integer. The irrational parts of ϕn\phi^n and ψn\psi^n cancel exactly upon subtraction.

Since ∣ψ∣<1|\psi| < 1, the term ψn\psi^n vanishes for large nn, and Fn≈ϕn5F_n \approx \frac{\phi^n}{\sqrt{5}}. This means the Fibonacci sequence grows exponentially at rate ϕ\phi — roughly 61.8%61.8\% growth per step.
Fibonacci numbers sit on the curve phi^n / sqrt 5: F10 = 55 and phi^10/sqrt5 = 55.004ny1234567891011102030405060708090y = φⁿ / √5F₁₀ = 55φ¹⁰/√5 ≈ 55.004: round it to get F₁₀the ψⁿ/√5 part is always under 1/2Fₙ is φⁿ/√5 rounded to the nearest whole number
Binet’s formula without the small term

The dots are F₁ to F₁₁; the dashed curve is φⁿ/√5 alone, with the ψⁿ term dropped. The dots sit on the curve: at n = 10 the curve gives 55.004 and F₁₀ = 55. The dropped part ψⁿ/√5 is smaller than 1/2 for every n, so rounding φⁿ/√5 to the nearest whole number gives Fₙ exactly.

Growth per step is the factor φ ≈ 1.618, the same φ the ratios converge to.

Identities and Properties

The Fibonacci sequence satisfies a network of algebraic identities. Several of the most fundamental are collected here.

Cassini's identity relates three consecutive Fibonacci numbers:

Cassini's Identity
Fn−1Fn+1−Fn2=(−1)nF_{n-1} F_{n+1} - F_n^2 = (-1)^n
Learn more about this formula: Cassini's Identity →

Fact: A chessboard that gains a square
Cut an 8×88 \times 8 chessboard into two triangles and two trapezoids, and the pieces seem to refill a 5×135 \times 13 rectangle of area 6565. The extra square is a thin gap along the diagonal, where the slopes 38\frac{3}{8} and 25\frac{2}{5} of the cut edges differ by only 140\frac{1}{40}.

The product of the neighbors minus the square of the middle term alternates between +1+1 and −1-1. For n=6n = 6: F5⋅F7−F62=5⋅13−82=65−64=1F_5 \cdot F_7 - F_6^2 = 5 \cdot 13 - 8^2 = 65 - 64 = 1.

Read as areas, the identity compares a rectangle with sides Fn−1F_{n-1} and Fn+1F_{n+1} against a square with side FnF_n. Their areas always differ by exactly one unit square: for n=6n = 6 the rectangle is larger, and for n=5n = 5 the square is, since 3⋅8−52=24−25=−13 \cdot 8 - 5^2 = 24 - 25 = -1. As nn grows, both shapes grow without bound, while the difference stays a single square.

Sum of the first n Fibonacci numbers:

Sum of Fibonacci Numbers
∑k=1nFk=Fn+2−1\sum_{k=1}^{n} F_k = F_{n+2} - 1
Learn more about this formula: Sum of Fibonacci Numbers →

The running total always lands one short of a Fibonacci number two positions ahead.

Sum of squares:

Sum of Squared Fibonacci Numbers
∑k=1nFk2=Fn⋅Fn+1\sum_{k=1}^{n} F_k^2 = F_n \cdot F_{n+1}
Learn more about this formula: Sum of Squared Fibonacci Numbers →

The sum of the first nn squared Fibonacci numbers equals the product of FnF_n and Fn+1F_{n+1}.

Divisibility: gcd⁡(Fm,Fn)=Fgcd⁡(m,n)\gcd(F_m, F_n) = F_{\gcd(m,n)}. The greatest common divisor of two Fibonacci numbers is itself a Fibonacci number, indexed by the GCD of the original indices. This connects the multiplicative structure of the Fibonacci sequence to the GCD of ordinary integers.

Fibonacci GCD Identity
gcd⁡(Fm,Fn)=Fgcd⁡(m,n)\gcd(F_m, F_n) = F_{\gcd(m,n)}
Learn more about this formula: Fibonacci GCD Identity →


Pitfall: Taking the GCD of the positions as the answer
Reading gcd⁡(F12,F18)\gcd(F_{12}, F_{18}) as gcd⁡(12,18)=6\gcd(12, 18) = 6 stops halfway: 66 does not even divide F18=2584F_{18} = 2584. The GCD of the positions is an index, so the answer is F6=8F_6 = 8, and indeed gcd⁡(144,2584)=8\gcd(144, 2584) = 8.

The identity works on two levels at once. On the left, the GCD is taken of two Fibonacci numbers, which are values in the sequence; on the right, it is taken of their positions mm and nn, and the result is used as a position again. For example, gcd⁡(F6,F9)=gcd⁡(8,34)=2=F3\gcd(F_6, F_9) = \gcd(8, 34) = 2 = F_3.

THEOREMZeckendorf's theorem
every positive integer can be represented as a sum of non-consecutive Fibonacci numbers, and this representation is unique.

For example, 30=21+8+1=F8+F6+F130 = 21 + 8 + 1 = F_8 + F_6 + F_1.
Identity Formula Example
Cassini's identity Fn−1 · Fn+1 − Fn² = (−1)n n = 6: 5·13 − 64 = 1
Sum of first n terms Σk=1n Fk = Fn+2 − 1 F₁+…+F₅ = 1+1+2+3+5 = 12 = F₇ − 1
Sum of squares Σk=1n Fk² = Fn · Fn+1 F₁²+…+F₅² = 1+1+4+9+25 = 40 = 5·8
Divisibility (GCD) gcd(Fm, Fn) = Fgcd(m, n) gcd(F₆, F₉) = gcd(8, 34) = 2 = F₃
Zeckendorf's theorem every positive integer is uniquely a sum of non-consecutive Fibonacci numbers 30 = 21 + 8 + 1 = F₈ + F₆ + F₁

Lucas Numbers

The Lucas sequence uses the same recurrence as Fibonacci but starts from different initial values:

Lucas Recurrence
L1=1,L2=3,Ln=Ln−1+Ln−2(n≥3)L_1 = 1, \quad L_2 = 3, \quad L_n = L_{n-1} + L_{n-2} \quad (n \geq 3)
Learn more about this formula: Lucas Recurrence →

The first terms are 1,3,4,7,11,18,29,47,76,123,…1, 3, 4, 7, 11, 18, 29, 47, 76, 123, \ldots. The growth rate is identical to Fibonacci — the ratio of consecutive Lucas numbers also converges to ϕ\phi — but the specific values differ.

The two sequences are tightly linked. The relationship Ln=Fn−1+Fn+1L_n = F_{n-1} + F_{n+1} expresses each Lucas number as the sum of the Fibonacci neighbors flanking it. For n=5n = 5: L5=F4+F6=3+8=11L_5 = F_4 + F_6 = 3 + 8 = 11.

Lucas-Fibonacci Relation
Ln=Fn−1+Fn+1L_n = F_{n-1} + F_{n+1}
Learn more about this formula: Lucas-Fibonacci Relation →

The Binet analogue for Lucas numbers is:

Lucas Binet Formula
Ln=ϕn+ψnL_n = \phi^n + \psi^n
Learn more about this formula: Lucas Binet Formula →


Fact: Primes leave a mark on Lucas numbers
For every prime pp, LpL_p is one more than a multiple of pp, as in L7=29=4⋅7+1L_7 = 29 = 4 \cdot 7 + 1: expanding (ϕ+ψ)p=1(\phi + \psi)^p = 1 leaves LpL_p plus middle terms that all carry the factor pp. The first composite number nn with the same property is 705705.

Where Fibonacci's formula subtracts powers of the two roots and divides by 5\sqrt{5}, Lucas's formula adds them directly. The complementary structure — one sequence from the difference, the other from the sum — is a consequence of both sequences satisfying the same characteristic equation.

The same holds for any pair of starting values: a sequence that follows the rule an=an−1+an−2a_n = a_{n-1} + a_{n-2} is always Aϕn+BψnA\phi^n + B\psi^n, with AA and BB fixed by its first two terms. Fibonacci has A=15A = \frac{1}{\sqrt{5}} and B=−15B = -\frac{1}{\sqrt{5}}; Lucas has A=B=1A = B = 1.

Whatever the starting values, the Fibonacci rule builds the sequence from ϕn\phi^n and ψn\psi^n; only the mix changes.

Fibonacci at a Glance

The page built the Fibonacci sequence from its recurrence, derived the golden ratio as the limit of consecutive-term ratios, obtained Binet's closed-form expression from the characteristic equation, surveyed five core identities, and introduced the Lucas-number companion sequence. The table below collects the structural facts — recurrence, roots, Binet form, growth rate, and the Lucas bridge — in one reference card.
Concept Statement Example / note
Recurrence Fn = Fn−1 + Fn−2 for n ≥ 3 F₃ = 1 + 1 = 2; F₄ = 1 + 2 = 3
Initial values F₁ = 1, F₂ = 1 (alternative convention starts F₀ = 0) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …
Characteristic equation x² = x + 1 — roots φ and ψ govern every closed form φ + ψ = 1, φ · ψ = −1
Golden ratio φ φ = (1 + √5) ⁄ 2 ≈ 1.6180; limit of Fn+1 ⁄ Fn F₆⁄F₅ = 1.6; F₁₀⁄F₉ ≈ 1.6176
Conjugate root ψ ψ = (1 − √5) ⁄ 2 ≈ −0.6180; |ψ| < 1 so ψn → 0 explains why ratios converge to φ, not ψ
Binet's formula Fn = (φn − ψn) ⁄ √5 — always an integer despite irrational inputs Fn ≈ φn ⁄ √5 for large n
Growth rate exponential at rate φ — roughly 61.8% growth per step faster than every arithmetic, slower than 2n
Lucas sequence same recurrence, L₁ = 1, L₂ = 3; Binet form Ln = φn + ψn 1, 3, 4, 7, 11, 18, 29, 47, 76, …
Fibonacci–Lucas bridge Ln = Fn−1 + Fn+1 L₅ = F₄ + F₆ = 3 + 8 = 11

Common Pitfalls

Four mistakes come up again and again with the Fibonacci sequence. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
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1Reading the list position as the index
When the sequence is written from F0=0F_0 = 0, the terms are counted from 11 as they stand in the list, and the kk-th entry is taken to be FkF_k. Every index then comes out one too high.
Wrong turn
Error here
In 0,1,1,2,3,5,8,13,21,34,55,…0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots the 10th entry is 3434, so F10=34F_{10} = 34.
Right path
The first entry of that list is F0F_0, so the kk-th entry is Fk−1F_{k-1}, and the 10th entry is F9=34F_9 = 34. The value F10F_{10} is the next one, 21+34=5521 + 34 = 55. It is the same in both conventions, since F0=0F_0 = 0 only adds a term in front of F1=F2=1F_1 = F_2 = 1.
Why it happensIn the list 1,1,2,3,…1, 1, 2, 3, \ldots that starts at F1F_1, entry number and index agree, and the habit of counting entries is carried over to a list that starts at F0F_0.
How to checkTest a small index. F3F_3 must be 22 in either convention, and Binet's formula with n=10n = 10 gives ϕ10−ψ105=55\frac{\phi^{10} - \psi^{10}}{\sqrt{5}} = 55.
2Dividing the earlier term by the later one
The ratio of consecutive terms is taken the other way up, smaller over larger, and the number it approaches is taken for the golden ratio.
Wrong turn
Error here
F10F11=5589≈0.618,so ϕ≈0.618\frac{F_{10}}{F_{11}} = \frac{55}{89} \approx 0.618, \quad \text{so } \phi \approx 0.618
Right path
The ratio is Fn+1Fn\frac{F_{n+1}}{F_n}, later over earlier.
F11F10=8955≈1.618,ϕ=1+52≈1.618\frac{F_{11}}{F_{10}} = \frac{89}{55} \approx 1.618, \quad \phi = \frac{1 + \sqrt{5}}{2} \approx 1.618
The reversed ratio approaches 1ϕ\frac{1}{\phi}, and dividing ϕ2=ϕ+1\phi^2 = \phi + 1 by ϕ\phi shows that 1ϕ=ϕ−1≈0.618\frac{1}{\phi} = \phi - 1 \approx 0.618.
Why it happens0.6180.618 and 1.6181.618 share their decimal digits, and both are quoted in connection with the golden ratio. The number 0.6180.618 also matches the size of ψ\psi, though ψ≈−0.618\psi \approx -0.618 is negative.
How to checkThe sequence grows, so a later term divided by an earlier one is greater than 11. The golden ratio must also satisfy ϕ2=ϕ+1\phi^2 = \phi + 1: 1.6182≈2.6181.618^2 \approx 2.618, while 0.6182≈0.3820.618^2 \approx 0.382, not 1.6181.618.
3Dropping the sign of the second root
In Binet's formula the second root is taken as 0.6180.618, positive, so ψn\psi^n gets the wrong sign whenever nn is odd.
Wrong turn
Error here
With ψ=5−12\psi = \frac{\sqrt{5} - 1}{2}, the cube is ψ3=5−2≈0.236\psi^3 = \sqrt{5} - 2 \approx 0.236, and with ϕ3=2+5\phi^3 = 2 + \sqrt{5},
F3=(2+5)−(5−2)5=45≈1.789F_3 = \frac{(2 + \sqrt{5}) - (\sqrt{5} - 2)}{\sqrt{5}} = \frac{4}{\sqrt{5}} \approx 1.789
Right path
ψ=1−52≈−0.618\psi = \frac{1 - \sqrt{5}}{2} \approx -0.618, so ψ3=2−5≈−0.236\psi^3 = 2 - \sqrt{5} \approx -0.236 and
F3=(2+5)−(2−5)5=255=2F_3 = \frac{(2 + \sqrt{5}) - (2 - \sqrt{5})}{\sqrt{5}} = \frac{2\sqrt{5}}{\sqrt{5}} = 2
Why it happens0.6180.618 is the familiar companion of ϕ\phi (it is 1ϕ\frac{1}{\phi}), and the minus sign of ψ\psi is lost when the two are confused. For even nn the sign makes no difference, so the slip shows only at odd nn.
How to checkBinet's formula must give a whole number. 45≈1.789\frac{4}{\sqrt{5}} \approx 1.789 is not one, while F3=2F_3 = 2 matches the list 1,1,2,…1, 1, 2, \ldots.
4Taking the GCD of the positions as the answer
In gcd⁡(Fm,Fn)=Fgcd⁡(m,n)\gcd(F_m, F_n) = F_{\gcd(m,n)}, the GCD of the indices is computed and given as the answer, without looking up the Fibonacci number at that position.
Start
gcd⁡(F12,F18)\gcd(F_{12}, F_{18})
Wrong turn
Error here
=gcd⁡(12,18)=6= \gcd(12, 18) = 6
Right path
gcd⁡(12,18)=6\gcd(12, 18) = 6 is a position, so gcd⁡(F12,F18)=F6=8\gcd(F_{12}, F_{18}) = F_6 = 8. Directly: F12=144=8⋅18F_{12} = 144 = 8 \cdot 18 and F18=2584=8⋅323F_{18} = 2584 = 8 \cdot 323, and 1818 and 323=17⋅19323 = 17 \cdot 19 share no factor, so gcd⁡(144,2584)=8\gcd(144, 2584) = 8.
Why it happensThe right-hand side contains a GCD of ordinary integers, and once it is worked out, the outer FF is overlooked. The identity mixes values and positions in one line.
How to checkThe answer must divide both numbers and must itself be a Fibonacci number. 66 does not divide 25842584, and 66 does not appear in 1,1,2,3,5,8,…1, 1, 2, 3, 5, 8, \ldots.
Common pitfalls with the Fibonacci sequence·/algebra/sequences/fibonacciLearn Math Class
Each slip uses the right pieces in the wrong place: the index is shifted by one, the ratio is turned upside down, the second root of x2=x+1x^2 = x + 1 loses its minus sign, and the GCD of the positions is taken for the GCD of the numbers. The checks share one move: test the claim against a value that is already known, such as F3=2F_3 = 2, ϕ≈1.618\phi \approx 1.618, or the list 1,1,2,3,5,8,…1, 1, 2, 3, 5, 8, \ldots

Fibonacci Sequence FAQ

Can every whole number be written as a sum of Fibonacci numbers?

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Yes, and in exactly one way if you forbid consecutive ones. That is Zeckendorf's theorem: every positive integer has a unique representation as a sum of non-consecutive Fibonacci numbers. For instance 30=21+8+130 = 21 + 8 + 1, which is F8+F6+F1F_8 + F_6 + F_1. Allowing consecutive terms would destroy the uniqueness.Read more →

Is the greatest common divisor of two Fibonacci numbers also a Fibonacci number?

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Always, and its index is the GCD of the original indices: gcd⁡(Fm,Fn)=Fgcd⁡(m,n)\gcd(F_m, F_n) = F_{\gcd(m,n)}. So the GCD of F12F_{12} and F18F_{18} is F6F_6, because gcd⁡(12,18)=6\gcd(12, 18) = 6. The identity ties the divisibility structure of the sequence directly to ordinary integer arithmetic on the positions.Read more →

What is Cassini's identity?

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It says that multiplying a Fibonacci number's two neighbours and subtracting the square of the number itself always gives ±1\pm 1: Fn−1Fn+1−Fn2=(−1)nF_{n-1}F_{n+1} - F_n^2 = (-1)^n. Taking n=6n = 6 gives 5⋅13−82=65−64=15 \cdot 13 - 8^2 = 65 - 64 = 1. The result alternates in sign with nn and never drifts further from zero.Read more →

Is the golden ratio 1.618 or 0.618?

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The golden ratio is ϕ=1+52≈1.618\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618, the limit of Fn+1Fn\frac{F_{n+1}}{F_n}, later term over earlier. The number 0.6180.618 is its reciprocal: dividing ϕ2=ϕ+1\phi^2 = \phi + 1 by ϕ\phi gives 1ϕ=ϕ−1\frac{1}{\phi} = \phi - 1, which some texts also call the golden ratio and which is what FnFn+1\frac{F_n}{F_{n+1}} approaches. The second root ψ≈−0.618\psi \approx -0.618 has the same digits with the opposite sign.Read more →

How can cutting up an 8×88 \times 8 square give an area of 6565?

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It cannot; the pieces only seem to fit. An 8×88 \times 8 board of area 6464, cut into two triangles and two trapezoids, appears to refill a 5×135 \times 13 rectangle of area 6565. The cut edges have slopes 38\frac{3}{8} and 25\frac{2}{5}, which differ by only 140\frac{1}{40}, so a thin gap of area 11 hides along the diagonal. The sides 5,8,135, 8, 13 are consecutive Fibonacci numbers, and the gap is Cassini's identity: 5⋅13−82=15 \cdot 13 - 8^2 = 1.Read more →

Do other starting numbers also lead to the golden ratio?

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Yes. Any sequence that follows the rule an=an−1+an−2a_n = a_{n-1} + a_{n-2} has the form Aϕn+BψnA\phi^n + B\psi^n, with AA and BB fixed by its first two terms. Since ∣ψ∣<1|\psi| < 1, the ψn\psi^n part dies away, so the ratio of consecutive terms tends to ϕ\phi whenever A≠0A \neq 0, which holds for every whole-number start except 0,00, 0. The Lucas numbers 1,3,4,7,11,…1, 3, 4, 7, 11, \ldots are one example.Read more →

Can Lucas numbers tell whether a number is prime?

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Only in one direction. For every prime pp, LpL_p is one more than a multiple of pp, as in L7=29=4⋅7+1L_7 = 29 = 4 \cdot 7 + 1, so any nn for which Ln−1L_n - 1 is not divisible by nn is certainly composite. Passing the check proves nothing, though: some composite numbers pass it too, the first being 705705.Read more →