A341291 - OEIS

A341291

a(n) = number of nodes of degree 3 or 4 that are at distance n from the origin in the graph of the Rick Kenyon tiling of the first quadrant by L and backwards-L tiles.

1, 2, 4, 6, 7, 9, 11, 14, 15, 17, 19, 20, 21, 23, 24, 28, 28, 31, 32, 35, 36, 39, 40, 44, 44, 46, 48, 49, 50, 52, 54, 57, 58, 60, 62, 64, 65, 66, 68, 71, 72, 74, 76, 77, 78, 80, 81, 84, 84, 87, 88, 90, 92, 94, 96, 99, 100, 102, 104, 105, 106, 108, 110, 113

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

There is a unique way to tile the first quadrant with L-tiles and backward-L tiles. Upside down-L's are not allowed.

Consider the graph formed by the 1-skeleton of the tiling, with nodes at points of degree 3 or 4. The sequence gives the number of nodes at distance 0 from the node at the origin.

Proof that this tiling is unique, from Jaap Scherphuis, Feb 19 2021 (Start):

Build the tiling row by row, from bottom to top, left to right.

When you fill a row, you must keep the row above tileable. Since all the tiles have a length-2 base, any unfilled part of a row with no spaces underneath it is only tileable if it has even length.

When you place a tile, its stalk protrudes into the row above, and only one choice of the tile makes the unfilled part to the left of the stalk have even length. (End)

It appears that the number of vertical infinite "cuts" in the wall that start at height n is given by the Jacobsthal sequence A001045. - Maurizio Paolini, Feb 09 2021

REFERENCES

Peter Winkler, Mathematical Puzzles, CRC Press, 2021; see p. lxx.

LINKS

Rémy Sigrist, Table of n, a(n) for n = 0..8191

Chaim Goodman-Strauss, Two-colored illustration of bottom six rows of tiling

Rémy Sigrist, Illustration of initial terms

Rémy Sigrist, Colored representation of the nodes at distance <= 511 from the origin (the color is a function of the distance)

Rémy Sigrist, C++ program for A341291

Index entries for coordination sequences

FORMULA

Label a grid point with a 1 if it is the left end-point of the base of a normal L tile, with a 2 if it is the right end-point of the base of a backwards L-tile (which we could call an R tile), and with a 0 otherwise.

Then it appears that above the line x = 3y/2, (x,y) has label 1 if x=3m, y=2n; 2 if x=3m, y=2n+1; and 0 otherwise. It would be nice to have a formula for the points below the line. - N. J. A. Sloane, Feb 11 2021

MATHEMATICA

(* Mathematica code from Chaim Goodman-Strauss for generating illustration, Feb 09 2021 *)

(* setting things up *)

Clear[L, R, l, r]; Protect[L, R, l, r];

(*the base of an L tile is LR; the top is either an l (over an L) or an r (over an R)*)

nextRow[lastRow_]:=(row ={}; While[Length[row]<width, row = Join[row, Switch[lastRow[[Length[row]+1]], L, {l, L, R}, R, {r}, _, {L, R}]]]; row);

draw[lr_, pos_, w_:1]:=Polygon[{w, 1}(#+pos+If[lr ===l, {.5, 0}, {-.5, 0}])&/@If[lr===l, {{-oin, -oin}, {oin, -oin}, {oin, -in}, {-in, -in}, {-in, oin}, {-oin, oin}, {-oin, -oin}}, {{-oin, -oin}, {oin, -oin}, {oin, oin}, {in, oin}, {in, -in}, {-oin, -in}, {-oin, -oin}}]];

drawTableau[tableau_]:=Graphics[{Blue, draw[l, Reverse@#]&/@Position[tableau, l], Red, draw[r, Reverse@#]&/@Position[tableau, r]}];

makedrawing:=(oin = 1-in; bottomRow =Join @@ Table[{L, R}, {width/2}]; drawTableau[NestList[nextRow, bottomRow, height]])

(* change these parameters as you please *)

width= 140;

height =12;

in=.1; (* the width of the "grout" between tiles *)

makedrawing

PROG

(C++) See Links section.

CROSSREFS

See A341292 for the number of tiles that are n tiles away from the initial tile.

Cf. A001045.