Polygonal Numbers

Summary: Polygonal numbers are the partial sums of arithmetical progressions starting at 1; each class (triangular, square, pentagonal, hexagonal, …) corresponds to a fixed common difference, and a single general formula covers all cases.

Sources: chapter-1.3.5, chapter-1.4.7, chapter-2.0.4

Last updated: 2026-05-05

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Origin

Take an arithmetical progression starting at 1 with common difference $d-1$ (where $d$ is the number of sides). The cumulative sums $1,1+t_2,1+t_2+t_3,\dots$ give the polygonal numbers for that type.

The main cases

Name	$d$ (diff)	$m$ (sides)	$n$-th term	Sequence
Triangular	$1$	$3$	$\frac{n(n+1)}{2}$	$1,3,6,10,15,\dots$
Square	$2$	$4$	$n^2$	$1,4,9,16,25,\dots$
Pentagonal	$3$	$5$	$\frac{n(3n-1)}{2}$	$1,5,12,22,35,\dots$
Hexagonal	$4$	$6$	$n(2n-1)$	$1,6,15,28,45,\dots$
Heptagonal	$5$	$7$	$\frac{n(5n-3)}{2}$	$1,7,18,34,55,\dots$

General formula

For an $m$-gon with side $n$:

$P(m,n)=\frac{(m-2)n^2-(m-4)n}{2}.$

Setting $m=2$ recovers the natural numbers ($n$ itself). Setting $m=3$ gives triangular numbers. Setting $m=4$ gives $n^2$, etc.

Notable relationships

• All hexagonal numbers are also triangular: the hexagonal sequence coincides with the odd-indexed triangular numbers $T_1,T_3,T_5,\dots$
• The sum of the first $n$ odd numbers equals $n^2$ (the $n$-th square number), linking square numbers to the progression $1,3,5,7,\dots$
• The $n$-th triangular number $T_n=\frac{n(n+1)}{2}$ also gives the sum of all natural numbers from 1 to $n$.

Extracting the root of a polygonal number (inverse problem)

Given a polygonal number $P$, finding its root $x$ requires solving a quadratic equation. Applying completing-the-square to each class gives:

Polygon	Root formula	Integrality condition
Triangle	$x=\frac{-1+\sqrt{8P+1}}{2}$	$8P+1$ a perfect square
Square	$x=\sqrt{P}$	$P$ a perfect square
Pentagon	$x=\frac{1+\sqrt{24P+1}}{6}$	$24P+1$ a perfect square
Hexagon	$x=\frac{1+\sqrt{8P+1}}{4}$	$8P+1$ a perfect square
Heptagon	$x=\frac{3+\sqrt{40P+9}}{10}$	$40P+9$ a perfect square
Octagon	$x=\frac{1+\sqrt{3P+1}}{3}$	$3P+1$ a perfect square

The triangular and hexagonal conditions are identical, confirming that every hexagonal number is also triangular. The general $n$-gonal root formula is:

$x=\frac{(n-4)+\sqrt{8(n-2)P+(n-4{)}^2}}{2(n-2)}$

See ch1.4.7-extraction-roots-polygonal-numbers for details and worked examples.

Triangular Numbers that are also Perfect Squares (§49)

A triangular number $T_x=x(x+1)/2$ is also a perfect square iff $2x^2+2x=(2T_x)\cdot 2$ is a square, i.e. $x^2+x$ must be a square. Setting $\sqrt{x^2+x}=mx/n$ and solving (Rule 2 of Chapter 2.0.4):

$x=\frac{2n^2}{m^2-2n^2}.$

$m$	$n$	$x$	$T_x$	$\sqrt{T_x}$
3	2	8	36	6
7	5	49	1225	35
17	12	288	41616	204

(source: chapter-2.0.4, §49). The sequence of such $x$ values (1, 8, 49, 288, …) is infinite.

Related pages

• arithmetical-progressions
• ch1.3.5-polygonal-numbers
• ch1.3.4-summation-arithmetical-progressions
• ch1.4.7-extraction-roots-polygonal-numbers
• completing-the-square
• quadratic-equations
• square-roots-and-irrational-numbers
• ch2.0.4-surd-rationalization