

A348227


Coordination sequence for Wilkinson's 123circle packing with respect to a circle of radius 1.


9



1, 4, 12, 20, 34, 38, 54, 54, 74, 70, 94, 86, 114, 102, 134, 118, 154, 134, 174, 150, 194, 166, 214, 182, 234, 198, 254, 214, 274, 230, 294, 246, 314, 262, 334, 278, 354, 294, 374, 310, 394, 326, 414, 342, 434, 358, 454, 374, 474, 390, 494, 406, 514, 422, 534, 438, 554
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OFFSET

0,2


COMMENTS

Wilkinson's 123circle packing (that is my name for it) is a packing of nonoverlapping circles in the plane, and can be seen in Figs. 1 and 1a. There are three sizes of circles: (a) radius 1, (b) radius 2, and (c) radius 3. Because 3^2 + 4^2 = 5^2, there is a rightangled triangle (shown in red in Fig. 3, "Tiling of plane ...") which when repeatedly translated and reflected builds the whole structure.
A convenient set of coordinates for the centers are: (a) radius 1: the points (8*i, 6*j), (b) radius 2: the points (8*i, 6*j+3), and (c) radius 3: the points (8*i+4, 6*j), where i and j take all integer values.
Let G denote the graph (Fig 1, Fig. 2) whose nodes correspond to the centers of the circles, with an edge for every pair of circles that touch. There are three types of nodes, corresponding to the circles of types (a), (b), and (c).
The cells in the planar graph G form a tiling of the plane by two kinds of triangles (345 and 556).
The coordination sequences for the three types of nodes in G are the present sequence, A348229, and A348231.
If we draw tangent lines through the points where any two circles touch, we obtain a tiling of the plane (Fig. 3) with three kinds of tiles: squares, irregular hexagons, and irregular octagons.
The boundary lines of the tiles form the cpq net (thanks to Davide M. Proserpio for this observation). See Fig. 4 and the RCSR link. This net is the dual graph to G. It has two kinds of nodes, whose coordination sequences are given in A348236 and A348237.


LINKS

Don Wilkinson, Letter to N. J. A. Sloane, Oct 12 1990. Among other things, this letter enclosed the illustration shown in the next link, and a copy of US Patent 4643307. [The handwritten notes at the top and bottom of the letter were added by me after the letter arrived.]


FORMULA

G.f. = (1+4*q+10*q^2+12*q^3+11*q^4+2*q^52*q^62*q^7)/(1q^2)^2. Discovered and proved using the "coloring book" method (see GoodmanStrauss & Sloane).
a(n) = 9*n + (n  2)*(1)^n  4 for n > 3.  Stefano Spezia, May 01 2022


EXAMPLE

We start at a black point (with a(0) = 1). It is joined to 4 neighbors (so a(1) = 4), which in turn are joined to 12 further points (so a(2) = 12), which are joined to 20 additional points (so a(2) = 20), and so on.


MATHEMATICA

A348227list[nmax_]:=PadRight[{1, 4, 12, 20}, nmax+1, Array[9#+(#2)(1)^#4&, nmax+1, 0]]; A348227list[100] (* Paolo Xausa, Aug 06 2023 *)


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



