Smaller of Thabit Pair is Tetrahedral

Theorem

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

$\ds m_1$

$=$

$\ds 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}$

$\ds m_2$

$=$

$\ds 2^n \times \paren {9 \times 2^{2 n - 1} - 1}$

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

We have that:

$\ds \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}$

$=$

$\ds 9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1$

$\ds $

$<$

$\ds 9 \times 2^{2 n - 1} - 1$

and so $m_1 < m_2$.

Then:

$\ds m_1$

$=$

$\ds 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}$

$\ds $

$=$

$\ds \frac {2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1} \times 6} 6$

$\ds $

$=$

$\ds \frac {\paren {3 \times 2^n} \times \paren {2 \times 3 \times 2^{n - 1} - 2} \times \paren {3 \times 2^n - 1} } 6$

$\ds $

$=$

$\ds \frac {\paren {3 \times 2^n} \times \paren {3 \times 2^n - 1} \times \paren {3 \times 2^n - 2} } 6$

$\ds $

$=$

$\ds \frac {k \times \paren {k + 1} \times \paren {k + 2} } 6$

where $k = 3 \times 2^n - 2$

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

$\blacksquare$