[seqfan] Re: Slicing a tetrahedron

[Veikko Pohjola]

Ed,
Your sequence A = {1,5,15,34,65,111,175,...}is not the sum of number of 
up-tetras, down-tetras and octas of all sizes, and thus it is not what is 
looked for here. It lists only the sum of number of those of the smallest 
size. Similar sequences are obtained for those of the second smallest size, 
third smallest, and so on. Summing up the number of all sizes of all the 
three types of polyhedra would give {1,6,20,50,104,193,329,526,...}. Of 
course we have corresponding sequences for each type separately.
It appears that this puzzle has not been treated fully by others. So I will 
consider submitting the sequences to OEIS.
Regards,
Veikko

----- Original Message ----- 
From: "Ed Jeffery" <lejeffery7 at gmail.com>
To: <seqfan at list.seqfan.eu>
Sent: Sunday, August 05, 2012 12:19 PM
Subject: [seqfan] Re: Slicing a tetrahedron


> 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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