[seqfan] Re: Slicing a tetrahedron

Ed Jeffery lejeffery7 at gmail.com
Sun Aug 5 11:19:00 CEST 2012


Yes I realized the error and spent hours drawing layers using a CAD program
to see what was going on. What I overlooked were the upright tetrahedra in
the middle of the layers as well as some inverted ones. There are
A000217(k) upright tetrahedra plus A000217(k-2) inverted ones plus
A000217(k-1) octahedra in the k-th layer, k>=3. Setting B(1) = 1 and B(2) =
4, the sequence B for totals in each layer is

B = {1,4,10,19,31,46,64,...}

from which the partial sums

A = {1,5,15,34,65,111,175,...},

as you stated, is the sequence Veikko was looking for. Sorry about that.

LEJ

> 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).



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