Smaller of Thabit Pair is Tetrahedral

Theorem
Let $\left({m_1, m_2}\right)$ be a Thabit pair such that $m_1 < m_2$.

Then $m_1$ is a tetrahedral number.

Proof
By Thabit's Rule:

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

We have that:

and so $m_1 < m_2$.

Then:

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