A298678 Start with the hexagonal tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of hexagonal tiles after n iterations.

1, 0, 7, 12, 73, 216, 919, 3204, 12409, 45408, 171271, 635580, 2379241, 8865000, 33113527, 123523572, 461111833, 1720661616, 6422058919, 23966525484, 89446140169, 333813840888, 1245817611991, 4649439829860, 17351975261881, 64758394108800, 241681735391047

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

For n > 0, a(n) is also the number of triangles with 6 markings after n iterations when starting with the hexagon.

a(n) is also the number of triangles with 6 markings after n iterations when starting with the triangle with 6 markings.

a(n) is also the number of hexagons after n iterations when starting with the triangle with 6 markings.

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. A298679, A298680, A298681, 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, 0, 1], i=0); while(1, print1(v[4], ", "); i++; if(i==n, break, v=substitute(v)))

PARI

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

Formula

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

13*a(n) = A077235(n) + 8*(-2)^n. - Bruno Berselli, Jan 25 2018

From Colin Barker, Jan 25 2018: (Start)

a(n) = (1/26)*((-1)^n*2^(4+n) + (5-2*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(5+2*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