Triangular number

A triangular number is a figurate number that represents the total number of objects arranged in the shape of an equilateral triangle, where the nth triangular number, denoted T_n, counts the dots forming a triangle with n dots along each side and is given by the formula T_n = \frac{n(n+1)}{2}.^[1]^[2] The sequence of triangular numbers begins with 0 (sometimes included), 1, 3, 6, 10, 15, 21, 28, 36, 45, and continues indefinitely, corresponding to the cumulative sums of the first natural numbers: T_1 = 1, T_2 = 1+2=3, T_3 = 1+2+3=6, and so on.^[2]^[3] This connection arises because each new layer added to the triangle contributes one more dot than the previous layer, mirroring the summation of consecutive integers.^[4] Triangular numbers are a specific case of polygonal numbers and can also be expressed using binomial coefficients as T_n = \binom{n+1}{2}, linking them to combinatorial mathematics.^[1] Known since antiquity, triangular numbers were studied by ancient Greek mathematicians, including the Pythagoreans, who associated them with geometric patterns and philosophical ideas about the natural world.^[5] The explicit multiplication formula for generating them was described by Diophantus in the third century AD, building on earlier work.^[6] Notable properties include their role in number theory, such as every positive integer being expressible as the difference of two consecutive triangular numbers, and connections to other sequences like squares and tetrahedrals.^[7] In modern contexts, they appear in computer science for algorithms involving summation and in puzzles like those solved by Carl Friedrich Gauss as a child.^[8]

Definition and Representation

Definition

A triangular number is a number obtained by adding the first n natural numbers, expressed as T_n = 1 + 2 + \dots + n, where n is a positive integer.^[1] This sequence arises from arranging objects in successively increasing rows to form a triangular pattern.^[9] The concept of triangular numbers traces its origins to ancient Greek mathematics, particularly among the Pythagoreans in the 6th century BCE, who studied them as part of figurate numbers—numerical representations of geometric shapes formed by arranging pebbles (calculi) or dots into patterns.^[7] These early mathematicians viewed such arrangements as embodying the harmony of numbers and geometry, with triangular numbers specifically evoking the simplest polygonal form.^[10] The first few triangular numbers are T_1 = 1, T_2 = 3, T_3 = 6, and T_4 = 10.^[1] Unlike other figurate numbers, which form shapes like squares or pentagons, triangular numbers distinctly represent the dots or objects in an equilateral triangle configuration.^[11] The table below lists the first 10 triangular numbers, along with simple text-based visualizations of their triangular arrays using asterisks (*) to represent dots:

n	T_n	Triangular Array Visualization
1	1	*
2	3	*
3	6	* * * *
4	10	* * * * * * * *
5	15	* * * * * * * * * * * * *
6	21	* * * * * * * * * * * * * * * * * * *
7	28	* * * * * * * * * * * * * * * * * * * * * * * * * *
8	36	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
9	45	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
10	55	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

Visual and Geometric Representation

Triangular numbers are constructed geometrically by arranging units, such as dots or objects, in successive rows that increase in length by one each time. The first triangular number, T_1, consists of a single unit; the second, T_2, adds a row of two units beneath it; the third, T_3, adds a row of three; and so on. This iterative process builds a structure where each subsequent triangular number incorporates all previous ones plus the new row, illustrating the cumulative nature of the sequence.^[5] The resulting arrangement forms the shape of an equilateral triangle, with the units aligned along the sides to create a symmetric, pointed figure. This visual form directly gives rise to the term "triangular numbers," as the pattern evokes the geometric properties of a triangle.^[5] To illustrate the growth, consider the following conceptual diagram of the first four triangular numbers using dots (•):

T₁ = 1
•

T₂ = 3
 •
• •

T₃ = 6
  •
 • •
• • •

T₄ = 10
   •
  • •
 • • •
• • • •
T₁ = 1
•

T₂ = 3
 •
• •

T₃ = 6
  •
 • •
• • •

T₄ = 10
   •
  • •
 • • •
• • • •

Such arrays highlight how the structure expands outward from a central point, maintaining triangular symmetry at each stage.^[12] In physical contexts, this geometric representation appears in everyday stacking arrangements, such as the arrangement of bowling pins in ten-pin bowling, which forms a triangle with 4 rows totaling 10 pins (T_4), or cannonballs piled into pyramidal stacks with triangular bases for stability.^[13]^[14] These models trace back to ancient discussions of figurate numbers, as described by Nicomachus of Gerasa in his Introduction to Arithmetic around 100 AD, where he explored the arrangement of objects into triangular forms to reveal numerical patterns.^[12]

Mathematical Formulation

Explicit Formulas

The nth triangular number T_n is given by the closed-form expression

T_n = \frac{n(n+1)}{2}.

^[1] This formula can be derived by considering the sum S = 1 + 2 + \cdots + n. Writing the sum in reverse order gives S = n + (n-1) + \cdots + 1. Adding these two identical sums term by term yields $2S = (n+1) + (n+1) + \cdots + (n+1) (with n terms), so $2S = n(n+1) and thus S = \frac{n(n+1)}{2}.^[1] An equivalent representation expresses the triangular number in terms of binomial coefficients:

T_n = \binom{n+1}{2}.

This follows directly from the definition of the binomial coefficient \binom{n+1}{2} = \frac{(n+1)n}{2}.^[15] The formula T_n = \frac{n(n+1)}{2} can be proved using mathematical induction. For the base case n=1, T_1 = 1 and \frac{1(1+1)}{2} = 1, which holds. Assume the statement is true for n=k, so T_k = \frac{k(k+1)}{2}. For n=k+1, T_{k+1} = T_k + (k+1) = \frac{k(k+1)}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = \frac{(k+1)(k+2)}{2}, completing the inductive step.^[16] To illustrate computation, consider n=5: T_5 = \frac{5 \times 6}{2} = 15, verified by the sum $1 + 2 + 3 + 4 + 5 = 15. Similarly, for n=10: T_{10} = \frac{10 \times 11}{2} = [55](/page/55), verified by $1 + 2 + \cdots + 10 = [55](/page/55).^[1]

Recursive and Generating Formulas

The recursive formula for triangular numbers provides an iterative method to construct the sequence by successively adding the next positive integer. It is defined with the base case T_0 = 0, and for n \geq 1, T_n = T_{n-1} + n.^[1] This approach mirrors the geometric interpretation of triangular numbers as the cumulative sum of the first n natural numbers, building each term incrementally from the previous one.^[17] To illustrate, starting from T_0 = 0:

T_1 = T_0 + 1 = 1
T_2 = T_1 + 2 = 3
T_3 = T_2 + 3 = 6
T_4 = T_3 + 4 = 10
T_5 = T_4 + 5 = 15
T_6 = T_5 + 6 = 21

This computation demonstrates how the formula generates the sequence step by step up to T_6 = 21.^[1] The ordinary generating function for the triangular numbers is G(x) = \sum_{n=1}^{\infty} T_n x^n = \frac{x}{(1 - x)^3}.^[1] This can be derived from the geometric series \sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} for |x| < 1. Differentiating yields \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1 - x)^2}, and multiplying by x gives \sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2}. Differentiating again produces \sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{1 + x}{(1 - x)^3}, so \sum_{n=1}^{\infty} n^2 x^n = \frac{x(1 + x)}{(1 - x)^3}. Since T_n = \frac{n(n + 1)}{2} = \frac{n^2 + n}{2}, the generating function follows as

G(x) = \frac{1}{2} \left( \frac{x(1 + x)}{(1 - x)^3} + \frac{x}{(1 - x)^2} \right) = \frac{x}{(1 - x)^3}. $$/5:_Additional_Topics/5.1:_Generating_Functions) In combinatorics, the hockey-stick identity relates to summing triangular numbers: $ \sum_{k=1}^m T_k = \binom{m+2}{3} $, which arises as a consequence of the identity $ \sum_{k=r}^n \binom{k}{r} = \binom{n+1}{r+1} $ applied to the binomial representation $ T_k = \binom{k+1}{2} $.[](https://mathworld.wolfram.com/ChristmasStockingTheorem.html) ## Fundamental Properties ### Arithmetic Properties Triangular numbers exhibit distinct parity patterns based on the index $n$. Specifically, $T_n$ is odd if $n \equiv 1 \pmod{4}$ or $n \equiv 2 \pmod{4}$, and even otherwise. To see this, consider the formula $T_n = \frac{n(n+1)}{2}$ modulo 2, which requires examining $n(n+1)$ modulo 4 since it is always even. For $n \equiv 0 \pmod{4}$, $n(n+1) \equiv 0 \pmod{4}$, so $T_n \equiv 0 \pmod{2}$. For $n \equiv 1 \pmod{4}$, $n(n+1) \equiv 2 \pmod{4}$, so $T_n \equiv 1 \pmod{2}$. For $n \equiv 2 \pmod{4}$, $n(n+1) \equiv 2 \pmod{4}$, so $T_n \equiv 1 \pmod{2}$. For $n \equiv 3 \pmod{4}$, $n(n+1) \equiv 0 \pmod{4}$, so $T_n \equiv 0 \pmod{2}$.[](https://math.stackexchange.com/questions/4818384/how-to-prove-that-the-triangular-numbers-follow-the-pattern-of-being-odd-odd-e) A key divisibility property is that every natural number can be expressed as the sum of at most three triangular numbers, allowing $T_0 = 0$ if necessary; this result, known as [Gauss's Eureka theorem](/page/Gauss's_Eureka_theorem), was noted in his diary in 1796. For example, 7 = $T_3 + T_1 + T_0 = 6 + 1 + 0$, and 4 = $T_2 + T_0 + T_0 = 3 + 0 + 0$.[](https://uva.theopenscholar.com/files/ken-ono/files/006_8.pdf) The product of two triangular numbers is not necessarily triangular. However, specific cases exist where it is, such as certain pairs where the indices satisfy particular relations; for instance, the product $T_1 \cdot T_2 = 1 \cdot 3 = 3 = T_2$ is triangular, though $T_2 \cdot T_3 = 3 \cdot 6 = 18$ is not. Products of consecutive triangular numbers like $T_n T_{n+1}$ are triangular only in isolated cases, such as $n=1$.[](https://www.tandfonline.com/doi/full/10.1080/0025570X.2024.2370070) Triangular numbers have asymptotic density zero in the natural numbers, meaning the proportion of triangular numbers up to $x$ tends to 0 as $x \to \infty$. This follows from their quadratic growth: the number of triangular numbers not exceeding $x$ is approximately $\sqrt{2x}$, so the density is on the order of $1/\sqrt{x}$.[](https://math.colgate.edu/~integers/uproc15/uproc15.pdf) ### Summation and Parity Characteristics The sum of the first $m$ triangular numbers is given by the formula

\sum_{k=1}^m T_k = \frac{m(m+1)(m+2)}{6}.

This expression, known as the $m$-th [tetrahedral number](/page/TetrahedralNumber), arises from substituting the explicit formula $T_k = \frac{k(k+1)}{2}$ into the sum, which simplifies using the known formulas for the sum of the first $m$ natural numbers $\sum k = \frac{m(m+1)}{2}$ and the sum of squares $\sum k^2 = \frac{m(m+1)(2m+1)}{6}$. The derivation telescopes through algebraic expansion and cancellation, yielding the compact binomial form $\binom{m+2}{3}$.[](https://mathworld.wolfram.com/TetrahedralNumber.html) Triangular numbers exhibit a repeating parity pattern every four terms: odd for $n \equiv 1, 2 \pmod{4}$, and even for $n \equiv 0, 3 \pmod{4}$. This alternating even-odd pattern in pairs was first observed by the Neo-Pythagorean mathematician [Nicomachus of Gerasa](/page/Nicomachus_of_Gerasa) in his *Introduction to Arithmetic* around 100 AD, where he explored the properties of [figurate numbers](/page/Figurate_number) and their numerical behaviors.[](http://staff.um.edu.mt/jmus1/Nicomachus.pdf) For a finer analysis, consider the residues of $T_n$ modulo 8, which reveal a periodic structure with period 16. The possible values are all residues from 0 to 7. The following table illustrates this for $n = 1$ to $16$: | $n$ | $T_n$ | $T_n \mod 8$ | |-------|---------|----------------| | 1 | 1 | 1 | | 2 | 3 | 3 | | 3 | 6 | 6 | | 4 | 10 | 2 | | 5 | 15 | 7 | | 6 | 21 | 5 | | 7 | 28 | 4 | | 8 | 36 | 4 | | 9 | 45 | 5 | | 10 | 55 | 7 | | 11 | 66 | 2 | | 12 | 78 | 6 | | 13 | 91 | 3 | | 14 | 105 | 1 | | 15 | 120 | 0 | | 16 | 136 | 0 | These residues can be derived by cases on $n \pmod{16}$, leveraging that $n(n+1)$ is always divisible by 2, and analyzing the higher powers of 2 in the numerator modulo 16 to account for the division. A key identity linking parity and squares is that $8T_n + 1$ is always a perfect square:

8T_n + 1 = (2n + 1)^2.

This follows directly from substituting $T_n = \frac{n(n+1)}{2}$ into the left side: $8 \cdot \frac{n(n+1)}{2} + 1 = 4n(n+1) + 1 = 4n^2 + 4n + 1 = (2n + 1)^2$. The result is an odd square, consistent with the fact that odd squares are congruent to 1 modulo 8. This identity connects to the [Pell equation](/page/Pell_equation) $x^2 - 2y^2 = \pm 1$, where solutions generate sequences involving triangular numbers through continued fraction expansions of $\sqrt{2}$, with $y$ often related to indices of triangular terms in fundamental solutions.[](https://mathworld.wolfram.com/TriangularNumber.html) ## Relations to Other Concepts ### Figurate and Polygonal Numbers Figurate numbers represent positive integers arranged in regular geometric patterns using equally spaced points, such as dots forming symmetric figures.[](https://mathworld.wolfram.com/FigurateNumber.html) Polygonal numbers form a specific subset of figurate numbers, where the points create the shape of a regular polygon with $k$ sides, known as $k$-gonal numbers; triangular numbers correspond to the 3-gonal case, depicting equilateral triangles.[](https://mathworld.wolfram.com/PolygonalNumber.html) The general formula for the $n$th $k$-gonal number is given by

P(k, n) = \frac{n \left[ (k-2)n - (k-4) \right]}{2},

which generates sequences for various polygons when $k \geq 3$.[](https://mathworld.wolfram.com/PolygonalNumber.html) For $k=3$, this simplifies to the triangular number formula

T_n = \frac{n(n+1)}{2},

confirming triangular numbers as the foundational polygonal series.[](https://mathworld.wolfram.com/TriangularNumber.html) This framework extends triangular constructions to higher polygons; for instance, square numbers ($k=4$) yield $P(4, n) = n^2$, while pentagonal numbers ($k=5$) are $P(5, n) = \frac{n(3n-1)}{2}$, with the first few terms being 1, 5, 12, 22, and 35.[](https://mathworld.wolfram.com/PentagonalNumber.html) In comparison, the initial triangular numbers are 1, 3, 6, 10, and 15, illustrating how polygonal layers build cumulatively from prior terms in a shared geometric progression.[](https://mathworld.wolfram.com/TriangularNumber.html) Notably, each pentagonal number equals one-third of a specific triangular number, as the $n$th pentagonal number $P(5, n) = T_{3n-1}/3$, linking the sequences through scaled triangular bases.[](https://mathworld.wolfram.com/PentagonalNumber.html) A related geometric construction involves centered polygonal numbers, which arrange layers around a central point rather than a vertex, offering conversions from standard polygonal forms. Centered triangular numbers, for example, represent triangles with a core dot surrounded by successive triangular rings, given by $\frac{3n^2 - 3n + 2}{2}$ for the $n$th term, with initial values 1, 4, 10, 19, and 31.[](https://mathworld.wolfram.com/CenteredTriangularNumber.html) This centered variant highlights how triangular numbers adapt to alternative figurate symmetries within the broader polygonal family.[](https://mathworld.wolfram.com/CenteredPolygonalNumber.html) ### Binomial and Polynomial Connections Triangular numbers are intimately connected to binomial coefficients through the identity $ T_n = \binom{n+1}{2} $, where $ \binom{\cdot}{\cdot} $ denotes the binomial coefficient.[](https://mathworld.wolfram.com/TriangularNumber.html) This equivalence arises because $ \binom{n+1}{2} = \frac{(n+1)n}{2} $, matching the standard formula for the $ n $-th triangular number. Combinatorially, $ \binom{n+1}{2} $ counts the number of ways to choose 2 distinct elements from a set of $ n+1 $ elements, providing an interpretation of $ T_n $ as the size of a complete graph $ K_{n+1} $ in terms of edges or the number of pairwise handshakes among $ n+1 $ people.[](https://mathworld.wolfram.com/TriangularNumber.html) The polynomial representation of triangular numbers further highlights their algebraic structure, as $ T_n $ evaluates the quadratic polynomial $ p(x) = \frac{x(x+1)}{2} $ at positive integers $ x = n $. This form underscores the second-degree nature of triangular numbers and facilitates connections to broader polynomial theory. Equivalently, $ T_n = \frac{(n+1)_2}{2!} $, where $ (n+1)_2 = (n+1)n $ is the falling factorial of order 2, linking triangular numbers to the basis of falling factorials in polynomial interpolation.[](https://mathworld.wolfram.com/TriangularNumber.html) Stirling numbers of the second kind appear in the change-of-basis between power and falling factorial bases, indirectly relating triangular numbers to these combinatorial objects via the quadratic polynomial's expansion, though the primary tie remains binomial. A key identity connecting triangular numbers to binomial sums is the hockey-stick identity: $ \sum_{i=r}^n \binom{i}{r} = \binom{n+1}{r+1} $ for integers $ n \geq r \geq 0 $.[](https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity) Specializing to $ r=2 $, this yields $ \sum_{i=2}^n \binom{i}{2} = \binom{n+1}{3} $. Since $ \binom{i}{2} = T_{i-1} $, the sum of the first $ n-1 $ triangular numbers is $ T_1 + T_2 + \cdots + T_{n-1} = \binom{n+1}{3} $, illustrating how binomial summation principles generate higher-order figurate numbers. This identity has a combinatorial proof by counting the ways to choose $ r+1 $ elements from $ n+1 $ with a distinguished largest element.[](https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity) ## Advanced Mathematical Aspects ### Triangular Roots The triangular root of a positive integer $k$ is defined as the positive real number $n$ satisfying $T_n = k$, where $T_n$ denotes the $n$th triangular number. By analogy with the square root, this inverse operation extends the concept to non-integer values of $n$ as well, though the focus here is on cases yielding integer $n$. To derive the explicit form, begin with the standard formula for the $n$th triangular number:

T_n = \frac{n(n+1)}{2} = k.

Multiplying both sides by 2 gives $n(n+1) = 2k$. Rearranging yields the quadratic equation $n^2 + n - 2k = 0$. Applying the quadratic formula, the positive root is

n = \frac{-1 + \sqrt{1 + 8k}}{2}.

This expression provides the exact triangular root for any $k > 0$.[](https://mathworld.wolfram.com/TriangularNumber.html) For the triangular root to be an [integer](/page/Integer) $n$, the [discriminant](/page/Discriminant) $1 + 8k$ must be a [perfect square](/page/Perfect_square), say $m^2$ where $m$ is a positive [odd](/page/Odd) [integer](/page/Integer). Then $n = (m - 1)/2$ is [integer](/page/Integer), and $k = T_n$. This [condition](/page/Condition) ensures $k$ is itself a triangular number. For example, when $k = 6$, $1 + 8 \cdot 6 = 49 = 7^2$, so $n = (7 - 1)/2 = 3$, and indeed $T_3 = 6$. Similarly, for $k = 10$, $1 + 80 = 81 = 9^2$, yielding $n = 4$.[](https://mathworld.wolfram.com/TriangularNumber.html) For large $k$, an approximation arises by expanding the square root: $\sqrt{1 + 8k} \approx \sqrt{8k} = 2\sqrt{2k}$, since the 1 becomes negligible. Substituting gives

n \approx \frac{-1 + 2\sqrt{2k}}{2} = \sqrt{2k} - \frac{1}{2}.