Equivalence of Definitions of Heptagonal Number

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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:

$H_0 = 0$

By vacuous summation:

$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4} = 0$


By definition of summation:

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

and so:

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

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:

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


Then:

$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4}$

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 $H_0 = 0 = \map P {7, 0}$.

Then:

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

Thus $\map P {7, n}$ and $H_n$ are generated by the same recurrence relation.

$\blacksquare$