[seqfan] Re: Slicing a tetrahedron

[William Rex Marshall]

On 5/08/2012 9:32 a.m., Ed Jeffery wrote:
> 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.

Incorrect, as 10 tetrahedra and 4 octahedra only fill 26/27 of the 
order-3 slicing. (The regular octahedron has a volume exactly four times 
that of the regular tetrahedron with the same edge length.) You also 
need to take account of A000292(k-2) inverted tetrahedra in the k-th 
layer (for k >= 3), so the total number of pieces is actually 1, 5, 15, 
34, 65, 111, 175, ... (A006003).