Smaller of Thabit Pair is Tetrahedral

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Theorem

Let $\tuple {m_1, m_2}$ be a Thabit pair such that $m_1 < m_2$.

Then $m_1$ is a tetrahedral number.


Proof

By Thabit's Rule:

\(\displaystyle m_1\) \(=\) \(\displaystyle 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}\)
\(\displaystyle m_2\) \(=\) \(\displaystyle 2^n \times \paren {9 \times 2^{2 n - 1} - 1}\)

for some $n \in \Z_{\ge 0}$.

We have that:

\(\displaystyle \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}\) \(=\) \(\displaystyle 9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1\)
\(\displaystyle \) \(<\) \(\displaystyle 9 \times 2^{2 n - 1} - 1\)

and so $m_1 < m_2$.

Then:

\(\displaystyle m_1\) \(=\) \(\displaystyle 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1} \times 6} 6\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {3 \times 2^n} \times \paren {2 \times 3 \times 2^{n - 1} - 2} \times \paren {3 \times 2^n - 1} } 6\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {3 \times 2^n} \times \paren {3 \times 2^n - 1} \times \paren {3 \times 2^n - 2} } 6\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {k \times \paren {k + 1} \times \paren {k + 2} } 6\) where $k = 3 \times 2^n - 2$

From Closed Form for Tetrahedral Numbers, this is the $k$th tetrahedral number where $k = 3 \times 2^n - 2$.

$\blacksquare$


Sources