Equivalence of Definitions of Heptagonal Number

Theorem

The following definitions of the concept of Heptagonal Number are equivalent:

Definition 1

$H_n = \begin{cases} 0 & : n = 0 \\ H_{n-1} + 5 n - 4 & : n > 0 \end{cases}$

Definition 2

$\displaystyle H_n = \sum_{i \mathop = 1}^n \left({5 i - 4}\right) = 1 + 6 + \cdots + \left({5 \left({n - 1}\right) - 4}\right) + \left({5 n - 4}\right)$

Definition 3

$\forall n \in \N: H_n = P \left({7, n}\right) = \begin{cases} 0 & : n = 0 \\ P \left({7, n - 1}\right) + 5 \left({n - 1}\right) + 1 & : n > 0 \end{cases}$

where $P \left({k, n}\right)$ denotes the $k$-gonal numbers.

Proof

Definition 1 implies Definition 2

Let $H_n$ be a heptagonal number by definition 1.

Let $n = 0$.

By definition:

$H_0 = 0$
$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \left({5 i - 4}\right) = 0$

By definition of summation:

 $\displaystyle H_{n - 1}$ $=$ $\displaystyle \sum_{i \mathop = 1}^{n - 1} \left({5 i - 4}\right)$ $\displaystyle$ $=$ $\displaystyle 1 + 6 + \cdots + 5 \left({n - 1}\right) + 4$

and so:

 $\displaystyle H_n$ $=$ $\displaystyle H_{n - 1} + 5 n - 4$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left({5 n - 4}\right)$ $\displaystyle$ $=$ $\displaystyle 1 + 6 + \cdots + 5 n - 4$

Thus $H_n$ is a heptagonal number by definition 2.

$\Box$

Definition 2 implies Definition 1

Let $H_n$ be a heptagonal number by definition 2.

Then:

 $\displaystyle H_n$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left({5 i - 4}\right)$ $\displaystyle$ $=$ $\displaystyle 1 + 6 + \cdots + \left({5 \left({n - 1}\right) - 4}\right) + \left({5 n - 4}\right)$ $\displaystyle$ $=$ $\displaystyle H_{n - 1} + 5 n - 4$

Then:

$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \left({5 i - 4}\right)$

is a vacuous summation and so:

$H_0 = 0$

Thus $H_n$ is a heptagonal number by definition 1.

$\Box$

Definition 1 equivalent to Definition 3

We have by definition that $H_0 = 0 = P \left({7, 0}\right)$.

Then:

 $\, \displaystyle \forall n \in \N_{>0}: \,$ $\displaystyle P \left({7, n}\right)$ $=$ $\displaystyle P \left({7, n - 1}\right) + \left({7 - 2}\right) \left({n - 1}\right) + 1$ $\displaystyle$ $=$ $\displaystyle P \left({7, n - 1}\right) + 5 \left({n - 1}\right) + 1$ $\displaystyle$ $=$ $\displaystyle P \left({7, n - 1}\right) + 5 n - 4$

Thus $P \left({7, n}\right)$ and $H_n$ are generated by the same recurrence relation.

$\blacksquare$