A298683 Start with the square tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of squares after n iterations.

1, 1, 1, 13, 37, 169, 577, 2269, 8245, 31225, 115633, 433357, 1613701, 6029641, 22488481, 83957053, 313274197, 1169270809, 4363546897, 16285441069, 60777168805, 226825331305, 846519962113, 3159262905757, 11790514883701, 44002830183481, 164220738741361

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 one more than the number of triangles with 4 markings after n iterations when starting with the square 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 (3,5,-9,2).

Crossrefs

Cf. A298678, A298679, A298680, A298681, A298682.

Programs

Mathematica

CoefficientList[Series[((1 - 2 x) (1 - 7 x^2))/((1 - x) (1 + 2 x) (1 - 4 x + x^2)), {x, 0, 26}], x] (* or *)
LinearRecurrence[{3, 5, -9, 2}, {1, 1, 1, 13}, 27] (* Michael De Vlieger, Jan 28 2018 *)
f[n_] := Simplify[(-13 + (-1)^(n + 1)*2^(2 + n) + (15 - 7 Sqrt[3])*(2 + Sqrt[3])^n + (2 - Sqrt[3])^n*(15 + 7 Sqrt[3]))/13]; Array[f, 28, 0] (* Robert G. Wilson v, Feb 26 2018 *)

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, 1, 0], i=0); while(1, print1(v[3], ", "); i++; if(i==n, break, v=substitute(v)))

PARI

Vec(((1-2*x)*(1-7*x^2))/((1-x)*(1+2*x)*(1-4*x+x^2)) + O(x^40)) \\ Colin Barker, Jan 25 2018

Formula

G.f.: ((1-2*x)*(1-7*x^2))/((1-x)*(1+2*x)*(1-4*x+x^2)). - Joerg Arndt, Jan 25 2018

From Colin Barker, Jan 25 2018: (Start)

a(n) = (1/13)*(-13 + (-1)^(1+n)*2^(2+n) + (15-7*sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(15+7*sqrt(3))).

a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.

(End)

a(n) = ((15 - 7*sqrt(3))*(2 + sqrt(3))^n + (2 - sqrt(3))^n*(15 + 7*sqrt(3)) - 4*(-2)^n)/13 - 1. - Bruno Berselli, Jan 25 2018