Equivalence of Definitions of Hexagonal Number

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Theorem

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

Definition 1

$H_n = \begin{cases}

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

Definition 2

$\ds H_n = \sum_{i \mathop = 1}^n \paren {4 \paren {i - 1} + 1} = 1 + 5 + \cdots + \paren {4 \paren {n - 2} + 1} + \paren {4 \paren {n - 1} + 1}$

Definition 3

$\forall n \in \N: H_n = \map P {6, n} = \begin{cases} 0 & : n = 0 \\ \map P {6, n - 1} + 4 \paren {n - 1} + 1 & : n > 0 \end{cases}$

where $\map P {k, n}$ denotes the $k$-gonal numbers.


Proof

Definition 1 implies Definition 2

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

Let $n = 0$.

By definition:

$H_0 = 0$

By vacuous summation:

$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {4 \paren {i - 1} + 1} = 0$


By definition of summation:

\(\ds H_{n-1}\) \(=\) \(\ds \sum_{i \mathop = 1}^{n - 1} \paren {4 \paren {i - 1} + 1}\)
\(\ds \) \(=\) \(\ds 1 + 5 + \cdots + 4 \paren {n - 2} + 1\)

and so:

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

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

$\Box$


Definition 2 implies Definition 1

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

Then:

\(\ds H_n\) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {4 \paren {i - 1} + 1}\)
\(\ds \) \(=\) \(\ds 1 + 5 + \cdots + \paren {4 \paren {n - 2} + 1} + \paren {4 \paren {n - 1} + 1}\)
\(\ds \) \(=\) \(\ds H_{n-1} + 4 \paren {n - 1} + 1\)


Then:

$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {4 \paren {i - 1} + 1}$

is a vacuous summation and so:

$H_0 = 0$

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

$\Box$


Definition 1 equivalent to Definition 3

We have by definition that $H_0 = 0 = \map P {6, 0}$.

Then:

\(\ds \forall n \in \N_{>0}: \, \) \(\ds \map P {6, n}\) \(=\) \(\ds \map P {6, n - 1} + \paren {6 - 2} \paren {n - 1} + 1\)
\(\ds \) \(=\) \(\ds \map P {6, n - 1} + 4 \paren {n - 1} + 1\)

Thus $\map P {6, n}$ and $H_n$ are generated by the same recurrence relation.

$\blacksquare$

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