Equivalence of Definitions of Octagonal Number

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Definition 1

$O_n = \begin{cases}

0 & : n = 0 \\ O_{n - 1} + 6 n - 5 & : n > 0 \end{cases}$

Definition 2

$\ds O_n = \sum_{i \mathop = 1}^n \paren {6 i - 5} = 1 + 7 + \cdots + \paren {6 \paren {n - 1} - 5} + \paren {6 n - 5}$

Definition 3

$\forall n \in \N: O_n = \map P {8, n} = \begin{cases}

0 & : n = 0 \\ \map P {8, n - 1} + 6 \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 $O_n$ be an octagonal number by definition 1.

Let $n = 0$.

By definition:

$O_0 = 0$

By vacuous summation:

$\ds O_0 = \sum_{i \mathop = 1}^0 \paren {6 i - 5} = 0$


By definition of summation:

\(\ds O_{n - 1}\) \(=\) \(\ds \sum_{i \mathop = 1}^{n - 1} \paren {6 i - 5}\)
\(\ds \) \(=\) \(\ds 1 + 7 + \cdots + 6 \paren {n - 1} - 5\)

and so:

\(\ds O_n\) \(=\) \(\ds O_{n - 1} + 6 n - 5\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {6 i - 5}\)
\(\ds \) \(=\) \(\ds 1 + 7 + \cdots + \paren {6 n - 5}\)

Thus $O_n$ is an octagonal number by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $O_n$ be an octagonal number by definition 2.

Then:

\(\ds O_n\) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {6 i - 5}\)
\(\ds \) \(=\) \(\ds 1 + 7 + \cdots + \paren {6 \paren {n - 1} - 5} + \paren {6 n - 5}\)
\(\ds \) \(=\) \(\ds O_{n - 1} + 6 n - 5\)


Then:

$\ds O_0 = \sum_{i \mathop = 1}^0 \paren {6 i - 5}$

is a vacuous summation and so:

$O_0 = 0$

Thus $O_n$ is an octagonal number by definition 1.

$\Box$


Definition 1 equivalent to Definition 3

We have by definition that $O_n = 0 = \map P {8, n}$.

Then:

\(\ds \forall n \in \N_{>0}: \, \) \(\ds \map P {8, n}\) \(=\) \(\ds \map P {8, n - 1} + \paren {8 - 2} \paren {n - 1} + 1\)
\(\ds \) \(=\) \(\ds \map P {8, n - 1} + 6 \paren {n - 1} + 1\)
\(\ds \) \(=\) \(\ds \map P {8, n - 1} + 6 n - 6 + 1\)
\(\ds \) \(=\) \(\ds \map P {8, n - 1} + 6 n - 5\)

Thus $\map P {8, n}$ and $O_n$ are generated by the same recurrence relation.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Octagonal_Number&oldid=544225"