bibkey: barker2018a299285
authors: N. J. A. Sloane, Colin Barker, Stefano Spezia
year: 2018
title: "OEIS A299285, coordination sequence for the tea 3D uniform tiling"
doi: null
url: https://oeis.org/A299285
claim: "Barker's generating function and recurrence for the tea coordination sequence, which Sloane marks as probably only conjectures, and Spezia's closed form a(n) = (31 - 3*(-1)^n + 126*n^2 + 4*A056594(n))/16 for n > 0, stated conditionally on them."
strata_touched:
  - D5/S3/StatisticalMechanics/TeaCoordinationSequence
license: citation-only
triage: anchor

OEIS A299285

Verified locator

Source: https://oeis.org/A299285

The entry was read through the plain-text endpoint https://oeis.org/search?q=id:A299285&fmt=text (revision 25, June 12, 2024). It was authored by N. J. A. Sloane (February 10, 2018); its first 20 terms were computed by D. M. Proserpio with ToposPro, and it links the Reticular Chemistry Structure Resource page of the tea net.

Statement

The entry is the coordination sequence of the tea uniform tiling of 3-space: $a(n)$ counts the vertices at graph distance exactly $n$ from a fixed vertex of its edge graph, with terms $1,10,33,73,128,199,285,388,506,640,789,955,\dots$. Its formulas are:

From Colin Barker, Feb 11 2018: (Start) G.f.: (1 + 8x + 14x^2 + 17x^3 + 14x^4 + 8x^5 + x^6) / ((1 - x)^3(1 + x)(1 + x^2)). a(n) = 2a(n-1) - a(n-2) + a(n-4) - 2a(n-5) + a(n-6) for n>6. (End) [I suspect Barker’s formulas only conjectures. - N. J. A. Sloane, Jun 12 2024] If the above formulas are true, then a(n) = (31 - 3(-1)^n + 126n^2 + 4A056594(n))/16 for n > 0. - Stefano Spezia, Jun 08 2024

Here A056594 is the period-four sequence $1,0,-1,0$.

Encoding and boundary

The repository presents the net on ${\mathbb{Z}}^3$: triangular lattices in the layers; between consecutive layers either a slab of tetrahedra and octahedra, whose orientation alternates between layers $k\equiv 0$ and $k\equiv 2(\mathrm{mod}4)$, or a slab of triangular prisms for odd $k$. It proves Spezia’s closed form, $8a(n)=63n^2+16,63n^2+17,63n^2+12,63n^2+17$ for $n\equiv 0,1,2,3(\mathrm{mod}4)$ and every $n\ge 1$, from the vertex $(0,0,0)$, through an exact distance formula (layer distance plus the distance to a hexagonal region determined by the slabs crossed). Barker’s generating function and recurrence follow from the closed form with $a(0)=1$ but are not stated separately; the partial sums A299286 and the other uniform tilings are outside this conclusion.