A298679 Start with the hexagonal tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of square tiles after n iterations.

0, 3, 6, 33, 102, 423, 1494, 5745, 21102, 79431, 295086, 1103985, 4114710, 15367143, 57329286, 213999153, 798569022, 2980473543, 11122931934, 41512040625, 154923657702, 578185735911, 2157812994486, 8053078824945, 30054477139470, 112164880064583

Offset: 0

Author

Felix Fröhlich, Jan 24 2018

Keywords

• nonn
• easy

Comments

The following substitution rules apply to the tiles:

triangle with 6 markings -> 1 hexagon

triangle with 4 markings -> 1 square, 2 triangles with 4 markings

square -> 1 square, 4 triangles with 6 markings

hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares

a(n) is also the number of triangles with 4 markings after n+1 iterations when starting with the hexagonal tile.

a(n) is also the number of square tiles after n+1 iterations when starting with the hexagonal tile.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• F. Gähler, Matching rules for quasicrystals: the composition-decomposition method, Journal of Non-Crystalline Solids, 153-154 (1993), 160-164.
• Tilings Encyclopedia, Shield
• Index entries for linear recurrences with constant coefficients, signature (2,7,-2).

Crossrefs

Cf. A298678, A298680, A298681, A298682, A298683.

Programs

Mathematica

CoefficientList[Series[ 3x/((1+2x)(1-4x+x^2)) ,{x,0,40}],x] (* or *) LinearRecurrence[{2,7,-2},{0,3,6},40] (* Harvey P. Dale, Mar 02 2022 *)

PARI

/* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 0, 0, 1], i=0); while(1, print1(v[3], ", "); i++; if(i==n, break, v=substitute(v)))

PARI

concat(0, Vec(3*x / ((1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ Colin Barker, Jan 25 2018

Formula

From Colin Barker, Jan 25 2018: (Start)

G.f.: 3*x / ((1 + 2*x)*(1 - 4*x + x^2)).

a(n) = (1/26)*(-3*(-1)^n*2^(2+n) + (6-5*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(6+5*sqrt(3))).

a(n) = 2*a(n-1) + 7*a(n-2) - 2*a(n-3) for n>2.

(End)

Extensions

More terms from Colin Barker, Jan 25 2018