A298681 Start with the square tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of triangles with 6 markings after n iterations.

0, 4, 4, 32, 80, 372, 1236, 4912, 17728, 67364, 248996, 934080, 3476400, 12993364, 48453364, 180907472, 675001760, 2519449092, 9402095556, 35090331232, 130956433168, 488740993844, 1823996357396, 6807266805360, 25405026124800, 94812927172324, 353846503607524

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

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, A298682, A298683.

Programs

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

PARI

concat(0, Vec(4*x*(1 - 2*x) / ((1 - 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.: 4*x*(1 - 2*x) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).

a(n) = (1/39)*(26 + (-1)^(1+n)*2^(5+n) + (3-9*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(3+9*sqrt(3))).

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

(End)

Extensions

More terms from Colin Barker, Jan 25 2018