Constructing a tetrahedron

The formulae for constructing the tetrahedron (thanks Michael Mendrin for providing these!) are as follows:

To make tetrahedra of linear size $n$ , you need

• $\frac{1}{6}n(n^2 - 1)$ octahedra
• $\frac{1}{3}n(n^2 + 2)$ tetrahedra

These equations can be derived as follows:

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Consider the triangular numbers, $\Delta_n = \frac{n(n+1)}{2}$

Also, lets consider $(O_n,T_n)$ for the number of octahedra and tetrehedra in a "great tetrahedra" of linear dimension, $n$ . Clearly for $n=1$ and $n=2$ we have:

• $(O_1,T_1) = (0,1)$
• $(O_2,T_2) = (1,4)$

(The second equation being the picture in the problem)

The thing to notice here, is that each time we add a row of tetrahedra and octahedra to construct the next "great tetrahedra" with one more linear dimension, we add $\Delta_{n-1}$ octahedra and $\Delta_n + \Delta_{n-2}$ tetrahedra.

So, for n>2

• $O_n = O_{n-1} + \Delta_{n-1}$
• $T_n = T_{n-1} + \Delta_{n} + \Delta_{n-2}$

These recurrence relations translate into:

• $O_n = \frac{1}{6}n(n^2 - 1)$
• $T_n = \frac{1}{3}n(n^2 + 2)$

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The first equation gives us $364$ octahedra for $n=13$ .

So, for $n=13$ the second equation gives us that there are $\boxed{741}$ tetrahedra.