ch1.3.5-polygonal-numbers

Chapter 1.3.5 — Of Figurate, or Polygonal Numbers

Summary: Euler derives the theory of polygonal numbers — triangular, square, pentagonal, hexagonal, and beyond — as partial sums of arithmetical progressions starting at 1, culminating in a single general formula for any $m$-gonal number.

Sources: chapter-1.3.5

Last updated: 2026-04-30

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Origin: partial sums of progressions (§425)

Polygonal numbers arise by forming the partial sums (1 term, 2 terms, 3 terms, …) of an arithmetical progression that begins at 1. The type of polygon is determined by the common difference.

Triangular numbers ($d=1$) (§426–429)

Starting from $1,2,3,4,5,\dots$ the cumulative sums are:

$1,3,6,10,15,21,28,36,45,55,66,\dots$

These are called triangular numbers because their units can always be arranged in a triangular array. The $n$-th triangular number equals the sum $1+2+\cdots +n$:

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

Since one of $n$ or $n+1$ is always even, this quotient is always an integer. For $n=100$: $T_{100}=5050$.

Square numbers ($d=2$) (§430–431)

Starting from $1,3,5,7,9,\dots$ (the odd numbers) the cumulative sums are:

$1,4,9,16,25,36,\dots =n^2.$

These are the ordinary square numbers, matching the result already established in ch1.3.4-summation-arithmetical-progressions (§422). Their point arrays form squares with $n$ points per side.

Pentagonal numbers ($d=3$) (§432–433)

Starting from $1,4,7,10,13,\dots$:

$1,5,12,22,35,51,70,92,117,\dots$

The $n$-th pentagonal number is:

$P_n=\frac{3n^2-n}{2}=\frac{n(3n-1)}{2}.$

Example: $n=7$ gives $P_7=70$; $n=100$ gives $P_{100}=14950$.

Hexagonal numbers ($d=4$) (§434–435)

Starting from $1,5,9,13,17,\dots$:

$1,6,15,28,45,66,91,120,153,\dots$

The $n$-th hexagonal number is:

$H_n=2n^2-n=n(2n-1).$

Key observation: every hexagonal number is also triangular. The hexagonal sequence is exactly the odd-indexed triangular numbers ($T_1,T_3,T_5,\dots$).

General polygonal formula (§436–437)

For any integer $m\ge 2$ (the number of sides of the polygon), with side $n$:

$\text{m-gonal number}=\frac{(m-2)n^2-(m-4)n}{2}.$

This unifies all cases:

Polygon	$m$	Formula
Bigon (natural numbers)	$2$	$n$
Triangle	$3$	$\frac{n(n+1)}{2}$
Square	$4$	$n^2$
Pentagon	$5$	$\frac{n(3n-1)}{2}$
Hexagon	$6$	$n(2n-1)$
Heptagon	$7$	$\frac{n(5n-3)}{2}$
Octagon	$8$	$n(3n-2)$
XII-gon	$12$	$n(5n-4)$
XX-gon	$20$	$n(9n-8)$
$m$-gon	$m$	$\frac{(m-2)n^2-(m-4)n}{2}$

Worked examples (§438–439)

• XXV-gonal number of side 36: formula gives $\frac{23\cdot 36^2-21\cdot 36}{2}=14526$.
• $365$-gonal number of side 12: $m=365$, $n=12$ gives $23970$.

Related pages

• polygonal-numbers
• ch1.3.4-summation-arithmetical-progressions
• arithmetical-progressions
• square-roots-and-irrational-numbers