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Drake Thomas and I have proposed a sequence A343909 to the On-Line Encyclopedia of Integer Sequences (OEIS), which counts "generalized polyforms": generalizations of free polyominoes (Tetris pieces), but with cells in the tetrahedral-octahedral honeycomb.

That is, take a subset of tetrahedral or octahedral cells in the honeycomb such that each cell is connected to every other cell by a chain of face-to-face connections. Consider two subsets to be the same if there is a rigid transformation that takes one to the other.

The ($0$-indexed) sequence begins: 1, 2, 1, 4, 9, 44, 195, 1186, 7385, 49444, 337504, ...

Which is to say that there is

  • 1 polyform with 0 cells (the empty polyform),
  • 2 polyforms with 1 cell (the tetrahedron and the octahedron),
  • 1 polyform with 2 cells (the tetrahedron attached to the octahedron by a face),
  • 4 polyforms with 3 cells, and
  • 9 with 4 cells, as shown below:
Animated GIF with three octahedra on the faces of a tetrahedron. Animated GIF with three tetrahedra on the faces of a octahedron. Animated GIF with three tetrahedra on the faces of a octahedron.
Animated GIF with three tetrahedra on the faces of a octahedron. Animated GIF with two tetrahedra and two octahedron connected face-to-face. Animated GIF with two tetrahedra and two octahedron connected face-to-face.
Animated GIF with two tetrahedra and two octahedron connected face-to-face. Animated GIF with two tetrahedra and two octahedron connected face-to-face. Animated GIF with two tetrahedra and two octahedron connected face-to-face.

Tetrahedra/octahedra breakdown

It's hard to compute large values of this sequence, but here is the breakdown of the number of octahedra and tetrahedra in $8$- and $9$-cell polyforms.:

$n = 8 \\ A343909(8) = 7385$

  • 1 octahedron, 7 tetrahedra: $1$
  • 2 octahedra, 6 tetrahedra: $285$
  • 3 octahedra, 5 tetrahedra: $3223$
  • 4 octahedra, 4 tetrahedra: $3440$
  • 5 octahedra, 3 tetrahedra: $432$
  • 6 octahedra, 2 tetrahedra: $4$

$n = 9 \\ A343909(9) = 49444$

  • 1 octahedron, 8 tetrahedra: $1$
  • 2 octahedra, 7 tetrahedra: $356$
  • 3 octahedra, 6 tetrahedra: $10853$
  • 4 octahedra, 5 tetrahedra: $27632$
  • 5 octahedra, 4 tetrahedra: $10141$
  • 6 octahedra, 3 tetrahedra: $459$
  • 7 octahedra, 2 tetrahedra: $2$

Questions

In these (and smaller) cases, the greatest number of polyforms occurs when the number of octahedra is $\lfloor (n-1)/2 \rfloor$. Based on the fact that the honeycomb is composed of alternating regular octahedra and tetrahedra in a ratio of 1:2, I would have expected the mode to be closer to one-third.

  • In the limit, is the mode asymptotic to $n/2$, $n/3$, or something else?
  • For a fixed number of cells, does the number of polyforms with $m$ octahedra have a known distribution in $m$?
  • What is the minimum number of octahedra than an $n$-cell polyform can have? The maximum number?
Glen Whitney
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Peter Kagey
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  • The first two questions strike me as much harder than the third. My intuition on the first is that the mode will remain roughly $n/2$, because you want to be able to "make as many turns as possible" to maximize the number of configurations. Incidentally, the two examples you give suggest a mode of $\lfloor n/2\rfloor$, not $\lfloor (n-1)/2\rfloor$ as you perhaps inadvertently wrote. – Glen Whitney Sep 08 '21 at 22:27
  • Oughtn't you perhaps submit a companion sequence to the OEIS that is the triangle (read by rows) where each row gives the individual numbers of polyoctets for a specific breakdown into octs and tets for a given total number of cells? Then A343909 would be the row sums of this triangle. I think it would make a valuable addition. – Glen Whitney Aug 17 '23 at 19:14
  • @GlenWhitney—just for you! :) https://oeis.org/history?seq=A365970 – Peter Kagey Sep 24 '23 at 02:52
  • Fingers crossed they accept it ;-) – Glen Whitney Sep 25 '23 at 00:22
  • Accepted sequence is here: https://oeis.org/A365970 – Peter Kagey Sep 25 '23 at 23:05

1 Answers1

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I will take a stab at the maximum number of octahedra: A sequence of octahedron-maximal polyoctets is obtained by starting with the unique 4-oct-1-tet pentoctet, and adding further 3-oct-1-tet tetroctet modules by their sole tetrahedral face in a zig-zag fashion along an axis which can extend indefinitely without any modules overlapping. Each member of this sequence is maximal because all tetrahedron faces are used and if there were any fewer tetrahedra, there would not be enough tetrahedron faces to attach all of the octahedra. Counting, the $k$th polyoctet in this sequence has $3k+1$ octahedra and $k$ tetrahedra, for $4k+1$ cells. For numbers of cells between 5,9,13, etc., the maximum is achieved simply by removing the proper number of "leaf" octahedra, since again if there were any fewer tetrahedra, it would be impossible to connect all octahedra.

Putting this all together, if there are $c$ cells, then at most $$c-\left\lfloor\frac{c-1}4\right\rfloor$$ of them are octahedra (the rest necessarily being tetrahedra).

I suspect the minimum number of octahedra should be accessible as well, although it's a bit trickier because of the fact that connecting an octahedron to one of the tetrahedra on a fully-surrounded octahedron actually connects to two of them simultaneously. If I get the details worked out I will update this answer.

Glen Whitney
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