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
- $\ds H_n = \sum_{i \mathop = 1}^n \paren {5 i - 4} = 1 + 6 + \cdots + \paren {5 \paren {n - 1} - 4} + \paren {5 n - 4}$
Definition 3
- $\forall n \in \N: H_n = \map P {7, n} = \begin{cases}
0 & : n = 0 \\ \map P {7, n - 1} + 5 \paren {n - 1} + 1 & : n > 0 \end{cases}$ where $\map P {k, n}$ denotes the $k$-gonal numbers.
Proof
Definition 1 implies Definition 2
Let $H_n$ be a heptagonal number by definition 1.
Let $n = 0$.
By definition:
- $H_0 = 0$
- $\ds H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4} = 0$
By definition of summation:
\(\ds H_{n - 1}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^{n - 1} \paren {5 i - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 6 + \cdots + 5 \paren {n - 1} + 4\) |
and so:
\(\ds H_n\) | \(=\) | \(\ds H_{n - 1} + 5 n - 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {5 n - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 6 + \cdots + 5 n - 4\) |
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:
\(\ds H_n\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {5 i - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 6 + \cdots + \paren {5 \paren {n - 1} - 4} + \paren {5 n - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds H_{n - 1} + 5 n - 4\) |
Then:
- $\ds H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4}$
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:
\(\ds \forall n \in \N_{>0}: \, \) | \(\ds \map P {7, n}\) | \(=\) | \(\ds \map P {7, n - 1} + \paren {7 - 2} \paren {n - 1} + 1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map P {7, n - 1} + 5 \paren {n - 1} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map P {7, n - 1} + 5 n - 4\) |
Thus $\map P {7, n}$ and $H_n$ are generated by the same recurrence relation.
$\blacksquare$