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%I #15 Jan 29 2018 06:07:40
%S 1,2,4,8,28,92,352,1280,4828,17900,67024,249680,932716,3479132,
%T 12987904,48464288,180885628,675045452,2519361712,9402270320,
%U 35089981708,130957132220,488739595744,1823999153600,6807261212956,25405037309612,94812904802704
%N Start with the triangle with 4 markings of the Shield tiling and recursively apply the substitution rule. a(n) is the number of triangles with 4 markings after n iterations.
%C The following substitution rules apply to the tiles:
%C triangle with 6 markings -> 1 hexagon
%C triangle with 4 markings -> 1 square, 2 triangles with 4 markings
%C square -> 1 square, 4 triangles with 6 markings
%C hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
%C a(n) is also one more than the number of squares after n iterations when starting with the triangle with 4 markings.
%H Colin Barker, <a href="/A298682/b298682.txt">Table of n, a(n) for n = 0..1000</a>
%H 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 Tilings Encyclopedia, <a href="https://tilings.math.uni-bielefeld.de/substitution/shield">Shield</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,5,-9,2).
%F From _Colin Barker_, Jan 25 2018: (Start)
%F G.f.: (1 + x)*(1 - 2*x - 5*x^2) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).
%F a(n) = (1/13)*(26 + (-2)^n + (2+sqrt(3))^n*(-7+5*sqrt(3)) - (2-sqrt(3))^n*(7+5*sqrt(3))).
%F a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
%F (End)
%o (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 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 terms(n) = my(v=[0, 1, 0, 0], i=0); while(1, print1(v[2], ", "); i++; if(i==n, break, v=substitute(v)))
%o (PARI) Vec((1 + x)*(1 - 2*x - 5*x^2) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)) + O(x^40)) \\ _Colin Barker_, Jan 25 2018
%Y Cf. A298678, A298679, A298680, A298681, A298683.
%K nonn,easy
%O 0,2
%A _Felix Fröhlich_, Jan 24 2018
%E More terms from _Colin Barker_, Jan 25 2018
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