Repunit in Base 9 is Triangular
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Theorem
Let $m$ be a repunit base $9$.
Then $m$ is a triangular number.
Proof
Let $m$ be a repunit base $9$ with $n$ digits.
We have:
\(\ds m\) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} 9^k\) | Basis Representation Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {9^n - 1} {9 - 1}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3^{2 n} - 1} 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {3^n - 1} \paren {3^n + 1} } 8\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {\paren {2 s + 1} - 1} \paren {\paren {2 s + 1} + 1} } 8\) | $3^n$ is odd, and so $3^n = 2 s + 1$ for some integer $s$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 s \paren {2 s + 2} } 8\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \dfrac {2 s} 2 \dfrac {2 s + 2} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {s \paren {s + 1} } 2\) |
The result follows from Closed Form for Triangular Numbers.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1,111,111,111,111,111,111$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1,111,111,111,111,111,111$