A pronic number is a number that is the product of two consecutive integers, that is, a number of the form ${\displaystyle n(n+1)}$.^[1] The study of these numbers dates back to Aristotle. They are also called oblong numbers, heteromecic numbers,^[2] or rectangular numbers;^[3] however, the term "rectangular number" has also been applied to the composite numbers.^[4][5]

The first few pronic numbers are:

0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462 … (sequence A002378 in the OEIS).

Letting ${\displaystyle P_{n}}$ denote the pronic number ${\displaystyle n(n+1)}$, we have ${\displaystyle P_{{-}n}=P_{n{-}1}}$. Therefore, in discussing pronic numbers, we may assume that ${\displaystyle n\geq 0}$ without loss of generality, a convention that is adopted in the following sections.

The pronic numbers were studied as figurate numbers alongside the triangular numbers and square numbers in Aristotle's Metaphysics,^[2] and their discovery has been attributed much earlier to the Pythagoreans.^[3] As a kind of figurate number, the pronic numbers are sometimes called oblong^[2] because they are analogous to polygonal numbers in this way:^[1]

1 × 2	2 × 3	3 × 4	4 × 5

The nth pronic number is the sum of the first n even integers, and as such is twice the nth triangular number^[1][2] and n more than the nth square number, as given by the alternative formula n² + n for pronic numbers. Hence the nth pronic number and the nth square number (the sum of the first n odd integers) form a superparticular ratio:

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

Due to this ratio, the nth pronic number is at a radius of n and n + 1 from a perfect square, and the nth perfect square is at a radius of n from a pronic number. The nth pronic number is also the difference between the odd square (2n + 1)² and the (n+1)st centered hexagonal number.

Since the number of off-diagonal entries in a square matrix is twice a triangular number, it is a pronic number.^[6]

The partial sum of the first n positive pronic numbers is twice the value of the nth tetrahedral number:

${\displaystyle \sum _{k=1}^{n}k(k+1)={\frac {n(n+1)(n+2)}{3}}=2T_{n}}$.

The sum of the reciprocals of the positive pronic numbers (excluding 0) is a telescoping series that sums to 1:^[7]

${\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i(i+1)}}={\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{20}}\cdots =1}$.

The partial sum of the first n terms in this series is^[7]

${\displaystyle \sum _{i=1}^{n}{\frac {1}{i(i+1)}}={\frac {n}{n+1}}}$.

The alternating sum of the reciprocals of the positive pronic numbers (excluding 0) is a convergent series:

${\displaystyle \sum _{i=1}^{\infty }{\frac {(-1)^{i+1}}{i(i+1)}}={\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{12}}-{\frac {1}{20}}\cdots =\log(4)-1}$.

Pronic numbers are even, and 2 is the only prime pronic number. It is also the only pronic number in the Fibonacci sequence and the only pronic Lucas number.^[8][9]

The arithmetic mean of two consecutive pronic numbers is a square number:

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

So there is a square between any two consecutive pronic numbers. It is unique, since

${\displaystyle n^{2}\leq n(n+1)<(n+1)^{2}<(n+1)(n+2)<(n+2)^{2}.}$

Another consequence of this chain of inequalities is the following property. If m is a pronic number, then the following holds:

${\displaystyle \lfloor {\sqrt {m}}\rfloor \cdot \lceil {\sqrt {m}}\rceil =m.}$

The fact that consecutive integers are coprime and that a pronic number is the product of two consecutive integers leads to a number of properties. Each distinct prime factor of a pronic number is present in only one of the factors n or n + 1. Thus a pronic number is squarefree if and only if n and n + 1 are also squarefree. The number of distinct prime factors of a pronic number is the sum of the number of distinct prime factors of n and n + 1.

If 25 is appended to the decimal representation of any pronic number, the result is a square number, the square of a number ending on 5; for example, 625 = 25² and 1225 = 35². This is so because

${\displaystyle 100n(n+1)+25=100n^{2}+100n+25=(10n+5)^{2}}$.

The difference between two consecutive unit fractions is the reciprocal of a pronic number:^[10]

${\displaystyle {\frac {1}{n}}-{\frac {1}{n+1}}={\frac {(n+1)-n}{n(n+1)}}={\frac {1}{n(n+1)}}}$