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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.

%I A298681 #13 Jan 29 2018 06:07:32
%S A298681 0,4,4,32,80,372,1236,4912,17728,67364,248996,934080,3476400,12993364,
%T A298681 48453364,180907472,675001760,2519449092,9402095556,35090331232,
%U A298681 130956433168,488740993844,1823996357396,6807266805360,25405026124800,94812927172324,353846503607524
%N 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.
%C A298681 The following substitution rules apply to the tiles:
%C A298681 triangle with 6 markings -> 1 hexagon
%C A298681 triangle with 4 markings -> 1 square, 2 triangles with 4 markings
%C A298681 square                   -> 1 square, 4 triangles with 6 markings
%C A298681 hexagon                  -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
%H A298681 Colin Barker, <a href="/A298681/b298681.txt">Table of n, a(n) for n = 0..1000</a>
%H A298681 F. Gähler, <a href="https://doi.org/10.1016/0022-3093(93)90335-U">Matching rules for quasicrystals: the composition-decomposition method</a>, Journal of Non-Crystalline Solids, 153-154 (1993), 160-164.
%H A298681 Tilings Encyclopedia, <a href="https://tilings.math.uni-bielefeld.de/substitution/shield">Shield</a>
%H A298681 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,5,-9,2).
%F A298681 From _Colin Barker_, Jan 25 2018: (Start)
%F A298681 G.f.: 4*x*(1 - 2*x) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).
%F A298681 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))).
%F A298681 a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
%F A298681 (End)
%o A298681 (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. */
%o A298681 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
%o A298681 terms(n) = my(v=[0, 0, 1, 0], i=0); while(1, print1(v[1], ", "); i++; if(i==n, break, v=substitute(v)))
%o A298681 (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
%Y A298681 Cf. A298678, A298679, A298680, A298682, A298683.
%K A298681 nonn,easy
%O A298681 0,2
%A A298681 _Felix Fröhlich_, Jan 24 2018
%E A298681 More terms from _Colin Barker_, Jan 25 2018