[seqfan] Re: Slicing a tetrahedron

[Ed Jeffery]

Veikko,

Let T denote the regular tetrahedron with unit edges. Let your slicing of T
corresponding to n be called an "n-slicing." An n-slicing produces n layers
(easiest when viewed parallel to the base of T) in which each layer
contains a number of congruent (again) regular tetrahedra t_n whose edges
are of length 1/n, and a number of congruent octahedra o_n each with eight
faces matching those of t_n. These pieces are easily counted as follows.

Let V_n(T) denote the total number of pieces (counting both types together)
after an n-slicing. For k in {1, ..., n}, the k-th layer contains
A000217(k) of the t_n and A000217(k-1) of the o_n, where A000217 =
{0,1,3,6,10,15,...} = Triangular numbers. Summing the layers for each piece
type then gives a total of A000292(n) of the t_n and A000292(n-1) of the
o_n, where A000292 = {0,1,4,10,20,35,...} = Tetrahedral numbers. Hence

V_n(T) = A000292(n) + A000292(n-1) = A000330(n),

where A000330 = {1,5,14,30,55,...} = Square pyramidal numbers.

Cf. http://oeis.org/A000217, http://oeis.org/A000292,
http://oeis.org/A000330.

LEJ

> Has this been treated somewhere?

> Slice a regular tetrahedron by planes parallel to each of the four faces
> such that the edges become divided into n equal pieces. What is the total
> number of tetrahedra and octahedra of all sizes thus formed as the function
> of n=1,2,3,...? (Not in OEIS)

> Regards,

> Veikko Pohjola