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Sequences



Definition and Notation
Sequence Notation
Explicit and Recursive Definitions
Arithmetic Sequences
Geometric Sequences
Harmonic Sequences
Fibonacci Numbers
Triangular Numbers
Square Numbers
Prime Numbers
Formula Summary
Common Pitfalls
Sequences FAQ



Terms, Indices, and Patterns

A sequence is an ordered list of numbers where each element occupies a definite position. The position is the index, the element at that position is the term, and the rule connecting indices to terms — whether a formula, a recurrence, or an observed pattern — is the sequence's defining characteristic. Some sequences arise from repeated addition, others from repeated multiplication, and others from rules that resist any simple algebraic description. This page introduces the common language and notation, then maps out the principal families: progressions, figurate numbers, recursive sequences, and primes.



Definition and Notation

What comes after 1, 2, 4, 8, 16?
Most people answer 3232: each number doubles the one before. But the same five numbers also count the regions a circle falls into when points on it are joined by chords — and that count continues with 3131.
Two rules agree on the first five positions and part at the sixth.
position nrule 1: double each timerule 2: regions of a circle cut by chords
111
222
344
488
51616
63231
The first five numbers cannot tell the two apart. What decides the sixth is the rule, not the list.
A sequence is fixed by its rule: for every position, the rule says which number stands there.
A sequence assigns a number to each positive integer. The input is the index nn, the output is the term ana_n. Writing a1,a2,a3,…a_1, a_2, a_3, \ldots lists the terms in order; the subscript tells you where each term sits.
The definition
A sequence is a function that assigns a number ana_n to each positive integer nn; ana_n is the term in position nn.
Formally, a sequence is a function f:N→Rf: \mathbb{N} \to \mathbb{R} — it takes a natural number and returns a real number. The function viewpoint makes precise what "ordered list" means: two sequences are equal exactly when an=bna_n = b_n for every index nn.
Two notations are standard. The brace form {an}n=1∞\{a_n\}_{n=1}^{\infty} names the entire sequence as a single object. The listing form a1,a2,a3,…a_1, a_2, a_3, \ldots displays individual terms. Both are used interchangeably throughout this section.
So doubling, an=2n−1a_n = 2^{n-1}, and the circle-region count are two different sequences, even though their first five terms are the same.
A sequence can stop after finitely many terms or go on forever — and "…" only works once the rule is known.
A sequence can be finite (a fixed number of terms, such as 3,7,11,153, 7, 11, 15) or infinite (continuing without end, such as 1,4,9,16,…1, 4, 9, 16, \ldots). The ellipsis signals that the pattern continues, though what "the pattern" is must be stated explicitly — the sequence 1,2,4,…1, 2, 4, \ldots could be powers of 22, or the start of something else entirely.
Fact: 1,2,4,8,161, 2, 4, 8, 16, then 3131
Mark nn points on a circle and join every pair by a chord, with no three chords meeting at one point. For n=1n = 1 to 55 the circle falls into 1,2,4,8,161, 2, 4, 8, 16 regions, but for n=6n = 6 into 3131: the count is (n4)+(n2)+1\binom{n}{4} + \binom{n}{2} + 1, not 2n−12^{n-1}.
A rule can be given as a formula for each term, or as a recipe that builds each term from earlier ones.
Doubling can be written an=2n−1a_n = 2^{n-1} or as "start at 11, then double". The two styles are compared in explicit and recursive definitions.
The rest of the page covers explicit and recursive definitions, arithmetic, geometric and harmonic sequences, and the Fibonacci, triangular, square and prime numbers.

Sequence Notation

Notation

Sequence Notation

A subscript for position, braces for the whole, an ellipsis for the rest — three ways of writing one ordered list. All are catalogued among the algebra symbols.
The subscripted term ana_n, the brace form and summation are set out in full at arithmetic sequence notation.
ana_n
a sub n — the nth term
The term in position nn: the input is the index nn, the output is the term ana_n. The subscript tells where each term sits.
CasesFormally a sequence is a function f:N→Rf: \mathbb{N} \to \mathbb{R}, so two sequences are equal exactly when an=bna_n = b_n for every index nn.
a1,a2,a3,…a_1, a_2, a_3, \ldots
a one, a two, a three, and so on
The listing form: individual terms displayed in order. The ellipsis signals that the pattern continues.
Do not confuseA pattern that states itself. 1,2,4,…1, 2, 4, \ldots could be powers of 22 or the start of something else entirely — the rule must be given explicitly.
{an}n=1∞\{a_n\}_{n=1}^{\infty}
the sequence a sub n, n from one to infinity
The brace form: names the entire sequence as a single object rather than listing its terms.
CasesUsed interchangeably with the listing form throughout this section.
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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.
SequencesThis page
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 sequence
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

Explicit and Recursive Definitions

There are two ways to specify a sequence. An explicit (closed-form) definition provides a formula that computes ana_n directly from nn. Given the index, you get the term in one step — no need to know any other terms. The sequence 1,4,9,16,25,…1, 4, 9, 16, 25, \ldots has the explicit definition an=n2a_n = n^2.

A recursive definition computes each term from one or more previous terms, together with enough initial values to get started. The same sequence of perfect squares can be defined recursively as a1=1a_1 = 1, an=an−1+2n−1a_n = a_{n-1} + 2n - 1. To find the 5050th term, you would need to compute all 4949 terms before it.

The trade-off is direct access versus structural transparency. An explicit formula lets you jump to any term instantly. A recursive rule reveals how the sequence builds on itself — each term emerges from the ones before it — but forces you to compute terms in order.

Fact: A simple rule no one can prove ends
Halve an even term; triple an odd term and add 11. Started at 2727, this recursion climbs as high as 92329232 and needs 111111 steps to reach 11, and no one has proved that every starting value reaches 11 at all.

Some sequences admit both forms. The arithmetic sequence 2,5,8,11,…2, 5, 8, 11, \ldots can be written explicitly as an=3n−1a_n = 3n - 1 or recursively as a1=2a_1 = 2, an=an−1+3a_n = a_{n-1} + 3. Others, like the sequence of prime numbers, have no known closed-form expression and resist explicit formulation entirely.

A formula reaches any term in one step; a recurrence reaches it only by building every term before it.

Arithmetic Sequences

An arithmetic sequence is built by adding the same fixed value — the common difference dd — to each term to produce the next. Starting from an initial term a1a_1, the sequence unfolds as a1,a1+d,a1+2d,a1+3d,…a_1, a_1 + d, a_1 + 2d, a_1 + 3d, \ldots, and the general term is:

General Term (Arithmetic Sequence)
an=a1+(n−1)da_n = a_1 + (n - 1)d
Learn more about this formula: General Term (Arithmetic Sequence) →


Pitfall: Counting n steps instead of n − 1
Writing an=a1+nda_n = a_1 + nd for 7,3,−1,−5,…7, 3, -1, -5, \ldots gives a4=7+4⋅(−4)=−9a_4 = 7 + 4 \cdot (-4) = -9, but the fourth term is −5-5. Place 44 is three steps from a1a_1: a4=7+3⋅(−4)=−5a_4 = 7 + 3 \cdot (-4) = -5.

The common difference governs direction and spacing. When d>0d > 0, the sequence increases; when d<0d < 0, it decreases; when d=0d = 0, every term is identical. The sequence 7,3,−1,−5,…7, 3, -1, -5, \ldots has d=−4d = -4, each term falling four units below the previous.

Arithmetic sequences produce the simplest summation formula in all of mathematics. The sum of the first nn terms — the arithmetic series — is:

Arithmetic Series Sum
Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)
Learn more about this formula: Arithmetic Series Sum →


The idea behind the formula is credited to Gauss: pair the first term with the last, the second with the second-to-last, and so on. Each pair sums to a1+ana_1 + a_n, and there are n2\frac{n}{2} such pairs.

To find a term, a sum or a missing value for a specific sequence, use the arithmetic sequence calculator.
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What the common difference decides
Each term of an arithmetic sequence is the previous one plus the same fixed value dd. The sign of dd governs the direction of the whole sequence.
Arithmetic sequence
an=a1+(n−1)da_n = a_1 + (n - 1)d
split by the sign of dd
↑
Case 1
d>0d > 0
Increases
each term is larger than the one before
↓
Case 2
d<0d < 0
Decreases
each term is smaller than the one before
e.g.7,3,−1,−5,…7, 3, -1, -5, \ldots with d=−4d = -4
=
Case 3
d=0d = 0
Constant
every term is identical
Takeaway
The size of dd sets the spacing; its sign sets the direction. The sequence 7,3,−1,−5,…7, 3, -1, -5, \ldots has d=−4d = -4, each term falling four units below the previous.
The sign of the common difference·/algebra/sequencesLearn Math Class

Geometric Sequences

A geometric sequence is built by multiplying each term by a fixed value — the common ratio rr — to produce the next. Starting from a1a_1, the terms are a1,a1r,a1r2,a1r3,…a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots, and the general term is:

General Term (Geometric Sequence)
an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}
Learn more about this formula: General Term (Geometric Sequence) →


Where arithmetic sequences grow (or shrink) by equal steps, geometric sequences grow (or shrink) by equal factors. The sequence 3,12,48,192,…3, 12, 48, 192, \ldots has r=4r = 4; the sequence 100,−50,25,−12.5,…100, -50, 25, -12.5, \ldots has r=−12r = -\frac{1}{2}, alternating in sign with each term.

The sum of a finite geometric series has its own closed form:

Geometric Sequences
Sn=a1⋅1−rn1−r(r≠1)S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad (r \neq 1)

RULE
When ∣r∣<1|r| < 1, the terms shrink toward zero fast enough that the infinite series converges to S=a11−rS = \frac{a_1}{1 - r}. When ∣r∣≥1|r| \geq 1, the terms do not diminish and the series diverges.

Pitfall: Summing a series that diverges
Applying a11−r\frac{a_1}{1 - r} to 3+12+48+192+⋯3 + 12 + 48 + 192 + \cdots gives 31−4=−1\frac{3}{1 - 4} = -1, a negative total for positive terms. With r=4r = 4 the partial sums 3,15,63,255,…3, 15, 63, 255, \ldots grow without bound, so there is no sum.

The two sum formulas are built from the same two numbers, a1a_1 and rr. The finite formula adds a fixed number of terms and holds for every ratio except r=1r = 1. The infinite formula is the value that the partial sums SnS_n settle on as nn grows, and it carries a condition on rr that the finite formula does not. Whether that condition holds is read from the ratio alone, before any sum is computed.
Arithmetic 2, 5, 8, ... adds 3; geometric 1, 2, 4, ... doubles and overtakes at n = 5ny123456748121620242832add 3: 2, 5, 8, 11, 14, 17double: 1, 2, 4, 8, 16, 32doubling passes adding 3 at n = 5Arithmetic grows by equal steps, geometric by equal factors
Equal steps against equal factors

The first six terms of two sequences, plotted against their index n. The arithmetic 2, 5, 8, 11, … adds 3 each time, so its dots climb a straight line. The geometric 1, 2, 4, 8, … doubles each time, so its dots start lower and bend upward. Doubling overtakes adding 3 at n = 5, where 16 passes 14, and the gap then widens with every term.

Any geometric sequence with a₁ > 0 and r > 1 eventually passes any arithmetic one, whatever its d.

Harmonic Sequences

A harmonic sequence is formed by taking the reciprocals of an arithmetic sequence. If the arithmetic progression has first term b1b_1 and common difference dd, the harmonic sequence is:

General Term (Harmonic Sequence)
an=1b1+(n−1)da_n = \frac{1}{b_1 + (n-1)d}
Learn more about this formula: General Term (Harmonic Sequence) →


The simplest example is 1,12,13,14,…1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots — the reciprocals of the natural numbers. The terms decrease toward zero, but they do so slowly. Unlike arithmetic and geometric sequences, harmonic sequences have no closed-form expression for their partial sums.

The harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} diverges, even though its terms approach zero. This stands in sharp contrast to the geometric series ∑12n\sum \frac{1}{2^n}, which converges. Approaching zero is necessary for convergence but not sufficient — the harmonic series is the most important illustration of this distinction.

THEOREMInequality of means
The harmonic sequence connects to the other two progressions through the inequality of means: for any set of positive numbers, the harmonic mean is never greater than the geometric mean, which is never greater than the arithmetic mean (H≤G≤AH \leq G \leq A).

Fibonacci Numbers

The Fibonacci sequence is defined by a two-term recurrence: each term is the sum of its two immediate predecessors.

Fibonacci Numbers
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)

This produces the sequence 1,1,2,3,5,8,13,21,34,55,89,144,…1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \ldots, where growth accelerates because each new term inherits the size of both its parents.

THEOREM
The ratio of consecutive Fibonacci numbers Fn+1Fn\frac{F_{n+1}}{F_n} converges to the golden ratio ϕ=1+52≈1.618\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618, a number that satisfies ϕ2=ϕ+1\phi^2 = \phi + 1.

Despite its recursive definition, the Fibonacci sequence has a closed-form expression known as Binet's formula:

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

where ψ=1−52\psi = \frac{1 - \sqrt{5}}{2}. The formula involves irrational numbers yet always yields an integer — a striking consequence of the algebraic relationship between ϕ\phi and ψ\psi.

Triangular Numbers

The triangular numbers count dots arranged in successively larger equilateral triangles: one dot in the first row, two in the second, three in the third, and so on. The nn-th triangular number is the sum of the first nn natural numbers:

Triangular Numbers
Tn=1+2+3+⋯+n=n(n+1)2T_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

The first terms are 1,3,6,10,15,21,28,36,…1, 3, 6, 10, 15, 21, 28, 36, \ldots. Recursively, Tn=Tn−1+nT_n = T_{n-1} + n — each triangular number is built from the previous one by adding a new row.

The formula Tn=n(n+1)2T_n = \frac{n(n+1)}{2} also equals the binomial coefficient (n+12)\binom{n+1}{2}, the number of ways to choose 22 items from n+1n + 1. This connection places triangular numbers at the intersection of summation, geometry, and combinatorics.

RULE
A key identity ties triangular and square numbers together: the sum of two consecutive triangular numbers is always a perfect square, Tn+Tn−1=n2T_n + T_{n-1} = n^2.

Geometrically, two triangles of consecutive sizes fit together to form a square.
11112113311464115101051C(5, 2) = 10 — 10 paths from the top
C(5, 2) = 10 is the fourth triangular number

The highlighted entry of Pascal’s triangle, C(5, 2), is reached by 10 downward paths, and 10 = 1 + 2 + 3 + 4 is a triangular number: the second diagonal of the triangle lists exactly the triangular numbers, because Tₙ = C(n + 1, 2). Counting dots in a triangle and choosing 2 objects from n + 1 are the same count. Follow the diagonal down on the binomial coefficient visualizer.

The square numbers, treated next, are sums of two consecutive triangular numbers.

Square Numbers

The square numbers are the perfect squares 1,4,9,16,25,36,…1, 4, 9, 16, 25, 36, \ldots, given by Sn=n2S_n = n^2. Each can be visualized as dots filling a square grid with nn rows and nn columns.

Recursively, the sequence satisfies Sn=Sn−1+(2n−1)S_n = S_{n-1} + (2n - 1). Each new square is built from the previous one by adding an L-shaped border of 2n−12n - 1 dots. This means every perfect square is a sum of consecutive odd numbers:

Square Numbers
n2=1+3+5+⋯+(2n−1)n^2 = 1 + 3 + 5 + \cdots + (2n - 1)

RULE
The difference between consecutive squares is always odd: Sn−Sn−1=2n−1S_n - S_{n-1} = 2n - 1.

This connects square numbers to arithmetic sequences, since the odd numbers 1,3,5,7,…1, 3, 5, 7, \ldots form an arithmetic progression with common difference 22.

The sum of the first nn squares has its own closed form:

Sum of Squares
∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
Learn more about this formula: Sum of Squares →


This formula appears throughout mathematics — in variance calculations, area approximations, and polynomial identities.

The square numbers calculator lists square numbers and tests whether a given number is a perfect square.

Each square grows from the one before by the next odd number, so the squares are running totals of the odd numbers.

Prime Numbers

A prime number is an integer greater than 11 whose only positive divisors are 11 and itself. The first primes are 2,3,5,7,11,13,17,19,23,29,…2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \ldots. The number 11 is excluded by convention — it has only one divisor, not two.

Unlike every other sequence on this page, the primes follow no algebraic pattern. There is no formula that generates the nn-th prime from nn, no common difference, no common ratio, and no recurrence relation. Each prime must be discovered individually, typically by verifying that no integer from 22 to p\sqrt{p} divides it.

Euclid proved that the primes are infinite: no finite list can contain them all. The argument assumes finitely many primes p1,p2,…,pkp_1, p_2, \ldots, p_k, forms the product N=p1p2⋯pk+1N = p_1 p_2 \cdots p_k + 1, and observes that NN is not divisible by any prime on the list — so either NN itself is prime or it has a prime factor missing from the list.

Pitfall: Reading N as always prime
Taking N=p1p2⋯pk+1N = p_1 p_2 \cdots p_k + 1 to be a new prime fails for the list 2,3,5,7,11,132, 3, 5, 7, 11, 13: N=30031=59⋅509N = 30031 = 59 \cdot 509. The proof only needs a prime factor of NN outside the list, here 5959 or 509509.

The construction works for any finite list of primes. For the list 2,3,52, 3, 5 it gives N=31N = 31, and for 2,3,5,72, 3, 5, 7 it gives N=211N = 211. In every case NN leaves remainder 11 when divided by each prime on the list, so none of them can be a factor of NN. That remainder is the only property of NN the argument uses.

Although primes never run out, they do thin out.

THEOREMPrime number theorem
The prime number theorem describes the rate: among integers near nn, roughly one in every ln⁡n\ln n is prime.

The primes become sparser as numbers grow, but they never vanish entirely.

The prime numbers calculator lists the first 10,000 primes, tests whether a number is prime and finds the nn-th prime.

Formula Summary

This summary collects the formulas of the page, one card per family of sequences: the general term on the left, and the sum or a second form on the right. Each card says how its terms are built and links to the section that explains it, and the buttons above the cards show one family at a time.

Arithmetic and geometric sequences each have a term formula and a sum formula; the geometric sum holds for every ratio except one. A harmonic sequence has only a term formula, since its partial sums have no closed form. The Fibonacci card sets the defining recurrence next to Binet's formula, and the last card holds the triangular and square numbers together with the sum of the first squares.
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Arithmetic2 formulas
Each term adds the same common difference dd to the one before; arithmetic sequences derives both formulas.
an=a1+(n−1)da_n = a_1 + (n - 1)d
The nn-th term: start at a1a_1 and add dd once for each of the n−1n - 1 steps.
Sn=n2(a1+an)\displaystyle S_n = \frac{n}{2}(a_1 + a_n)
The sum of the first nn terms: the number of terms times the average of the first and the last.
Geometric2 formulas
Each term multiplies the one before by the same common ratio rr; the formulas come from geometric sequences.
an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}
The nn-th term: start at a1a_1 and multiply by rr once for each of the n−1n - 1 steps.
Sn=a1⋅1−rn1−r(r≠1)\displaystyle S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad (r \neq 1)
The sum of the first nn terms, for every ratio except r=1r = 1.
Harmonic1 formula
The reciprocals of an arithmetic sequence, as harmonic sequences shows.
an=1b1+(n−1)d\displaystyle a_n = \frac{1}{b_1 + (n-1)d}
One over the nn-th term of the arithmetic sequence with first term b1b_1 and difference dd. There is no sum formula: the partial sums have no closed form.
Fibonacci2 formulas
Each term is the sum of the two before it; Fibonacci numbers gives both forms.
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)
The definition: two starting ones, then each term adds its two predecessors.
Fn=ϕn−ψn5\displaystyle F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}
Binet's formula gives FnF_n directly from the golden ratio ϕ\phi and ψ=1−52\psi = \frac{1 - \sqrt{5}}{2}, and always returns a whole number.
Triangular and square3 formulas
Numbers that count dots in triangles and squares, from triangular numbers and square numbers.
Tn=1+2+3+⋯+n=n(n+1)2\displaystyle T_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
The nn-th triangular number is the sum of the first nn natural numbers.
n2=1+3+5+⋯+(2n−1)n^2 = 1 + 3 + 5 + \cdots + (2n - 1)
Every perfect square is a sum of consecutive odd numbers.
∑k=1nk2=n(n+1)(2n+1)6\displaystyle \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
The sum of the first nn square numbers.
Sequence formulas·All algebra formulasLearn Math Class
The cards repeat one pattern. Once the first term and the rule that builds the next one are known, the general term gives any term straight from its position, without listing the terms before it, and where a sum formula exists it adds up a whole stretch of terms in one step.

Common Pitfalls

Three mistakes come up often with the sequences on this page. 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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1Counting n steps instead of n − 1
In the general term of an arithmetic sequence, dd is multiplied by the position nn instead of by n−1n - 1, the number of steps from the first term.
Start
For 7,3,−1,−5,…7, 3, -1, -5, \ldots with a1=7a_1 = 7 and d=−4d = -4,
Wrong turn
Error here
a4=7+4⋅(−4)=−9a_4 = 7 + 4 \cdot (-4) = -9
Right path
From place 11 to place 44 there are 33 steps.
a4=7+(4−1)⋅(−4)=7−12=−5a_4 = 7 + (4 - 1) \cdot (-4) = 7 - 12 = -5
This matches the fourth term of the listed sequence.
Why it happensThe formula is remembered as "first term plus nn times dd", and the fact that a1a_1 already occupies the first place, before any dd is added, gets lost.
How to checkPut n=1n = 1 into the formula; it must return a1a_1. The wrong version gives 7+1⋅(−4)=37 + 1 \cdot (-4) = 3, which is the second term, while 7+0⋅(−4)=77 + 0 \cdot (-4) = 7.
2Summing a series that diverges
The infinite-sum formula S=a11−rS = \frac{a_1}{1 - r} is applied to a geometric series whose ratio has size 11 or more.
Start
For 3+12+48+192+⋯3 + 12 + 48 + 192 + \cdots with a1=3a_1 = 3 and r=4r = 4,
Wrong turn
Error here
S=31−4=3−3=−1S = \frac{3}{1 - 4} = \frac{3}{-3} = -1
Right path
∣r∣=4≥1|r| = 4 \geq 1, so the terms do not shrink. The partial sums are 3,15,63,255,…3, 15, 63, 255, \ldots, and they grow without bound. The series diverges and has no sum.
Why it happensThe formula needs only a1a_1 and rr and returns a number for every ratio except r=1r = 1. The condition ∣r∣<1|r| < 1 is written beside it, not inside it, and is easy to leave behind.
How to checkCompare the answer with the first few partial sums. A series of positive terms cannot add up to a negative number: here every partial sum is positive and larger than the one before, so the sums cannot be approaching −1-1. Checking ∣r∣<1|r| < 1 first avoids the problem.
3Reading N as always prime
In Euclid's proof, the number N=p1p2⋯pk+1N = p_1 p_2 \cdots p_k + 1 is taken to be a new prime, as if the construction always produced primes.
Wrong turn
Error here
The lists 22, then 2,32, 3, then 2,3,52, 3, 5, then 2,3,5,72, 3, 5, 7, then 2,3,5,7,112, 3, 5, 7, 11 give N=3,7,31,211,2311N = 3, 7, 31, 211, 2311, all prime, so the next list should give a prime too:
N=2⋅3⋅5⋅7⋅11⋅13+1=30031 is primeN = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 30031 \text{ is prime}
Right path
30031=59⋅50930031 = 59 \cdot 509, so NN is not prime. Neither 5959 nor 509509 is on the list 2,3,5,7,11,132, 3, 5, 7, 11, 13, so the list still misses a prime, which is all the proof claims.
Why it happensThe first five cases are all prime, and the proof is often summarised as "multiply the primes and add one to get a new prime", which drops the second branch of the argument.
How to checkThe proof ends in two cases, "NN is prime" or "NN has a prime factor missing from the list". Either way a prime is missing, and the argument never needs to know which case holds. Dividing 3003130031 by 5959 gives exactly 509509.
Common pitfalls with sequences·/algebra/sequencesLearn Math Class
The three slips come from three different topics on this page, but each one takes a formula or an argument beyond what it says. The nn-th term formula counts n−1n - 1 steps from the first term, the sum a11−r\frac{a_1}{1 - r} holds only when ∣r∣<1|r| < 1, and Euclid's proof shows that a list of primes always misses one, not that NN itself is prime. Putting n=1n = 1 into the formula, comparing with the first partial sums, and factoring 3003130031 settle the three cases.

Sequences FAQ

What comes after 1,2,4,8,161, 2, 4, 8, 16?

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It depends on the rule, not on the list. Doubling gives 3232, from an=2n−1a_n = 2^{n-1}. But the same five numbers count the regions a circle falls into when nn points on it are joined by every chord, with no three chords meeting at one point, and that count continues with 3131. Its rule is (n4)+(n2)+1\binom{n}{4} + \binom{n}{2} + 1, which agrees with 2n−12^{n-1} for n=1n = 1 to 55 and then falls behind. A few terms never fix a sequence; only its rule does.Read more →

Does every starting number reach 11 under the halve-or-triple-plus-one rule?

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No one knows. The rule halves an even term and turns an odd term tt into 3t+13t + 1. Every starting value tried so far reaches 11, but no proof exists that all of them do; the question is known as the Collatz conjecture. Even small starts can wander far: from 2727 the sequence climbs as high as 92329232 and needs 111111 steps to reach 11. A recursive rule this short can still be beyond anyone's power to predict.Read more →

Does a geometric sequence always overtake an arithmetic one?

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Yes, when the geometric sequence has a1>0a_1 > 0 and r>1r > 1. An arithmetic sequence moves by equal steps dd, while a geometric one moves by equal factors, so its steps grow with every term and eventually outrun any fixed dd. Compare 2,5,8,11,…2, 5, 8, 11, \ldots, adding 33, with 1,2,4,8,…1, 2, 4, 8, \ldots, doubling: at n=4n = 4 the terms are 1111 and 88, but at n=5n = 5 they are 1414 and 1616, and from then on the gap only widens.Read more →

Which other sequences are built from arithmetic sequences?

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At least three on this page. A harmonic sequence is the reciprocals of an arithmetic one: 1,12,13,…1, \frac{1}{2}, \frac{1}{3}, \ldots comes from 1,2,3,…1, 2, 3, \ldots. The triangular numbers are running totals of 1,2,3,…1, 2, 3, \ldots (common difference 11): Tn=1+2+⋯+nT_n = 1 + 2 + \cdots + n. The square numbers are running totals of the odd numbers 1,3,5,…1, 3, 5, \ldots (common difference 22): n2=1+3+⋯+(2n−1)n^2 = 1 + 3 + \cdots + (2n - 1).Read more →

Is there a formula for the nnth prime?

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No practical one is known. Unlike the other sequences on this page, the primes have no common difference, no common ratio and no recurrence, so each prime is found on its own, usually by checking that no integer from 22 up to p\sqrt{p} divides it. What is known is how they thin out: by the prime number theorem, roughly one integer in every ln⁡n\ln n near nn is prime, so about one in 1414 near a million.Read more →