Polygonal number ($P_{s,n}$)

For integers $s\ge 3$ and $n\ge 1$, the $n$th $s$-gonal number isThese figurate numbers begin at one, and their successive increments form an arithmetic progression:Summing these increments proves the formula. For $s=3,4,5,6$, it gives triangular numbers, square numbers, pentagonal numbers and hexagonal numbers. For example, the first three pentagonal numbers are $1,5,12$, whose successive increments are four and seven. Increasing the number of polygon sides changes the step between increments while keeping this common quadratic structure.

The polygonal number $P_{8,n}=n(3n-2)$ for a positive integer $n$.

The polygonal number $P_{7,n}=n(5n-3)/2$ for a positive integer $n$.

The $n$th hexagonal number is $H_n=n(2n-1)$ for a positive integer $n$. Since $H_n=T_{2n-1}$, every hexagonal number is also a triangular number.

The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

A value $k(3k-1)/2$ with $k$ any integer, including negative values and zero. For $k=\pm 1,\pm 2,\dots$, positive values occur as pairs $j(3j-1)/2$ and $j(3j+1)/2$. These are the indices in the Euler pentagonal identity.