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 A298679 Start with the hexagonal tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of square tiles after n iterations. 6
 0, 3, 6, 33, 102, 423, 1494, 5745, 21102, 79431, 295086, 1103985, 4114710, 15367143, 57329286, 213999153, 798569022, 2980473543, 11122931934, 41512040625, 154923657702, 578185735911, 2157812994486, 8053078824945, 30054477139470, 112164880064583 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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 the number of triangles with 4 markings after n+1 iterations when starting with the hexagonal tile. a(n) is also the number of square tiles after n+1 iterations when starting with the hexagonal 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 (2,7,-2). FORMULA From Colin Barker, Jan 25 2018: (Start) G.f.: 3*x / ((1 + 2*x)*(1 - 4*x + x^2)). a(n) = (1/26)*(-3*(-1)^n*2^(2+n) + (6-5*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(6+5*sqrt(3))). a(n) = 2*a(n-1) + 7*a(n-2) - 2*a(n-3) for n>2. (End) MATHEMATICA CoefficientList[Series[ 3x/((1+2x)(1-4x+x^2)) , {x, 0, 40}], x] (* or *) LinearRecurrence[{2, 7, -2}, {0, 3, 6}, 40] (* Harvey P. Dale, Mar 02 2022 *) PROG (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 > 0, w++; v--); while(v > 0, w++; w=w+2; v--); while(v > 0, w++; w=w+4; v--); while(v > 0, w=w+7; w=w+3; w=w+3; v--)); w terms(n) = my(v=[0, 0, 0, 1], i=0); while(1, print1(v, ", "); i++; if(i==n, break, v=substitute(v))) (PARI) concat(0, Vec(3*x / ((1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ Colin Barker, Jan 25 2018 CROSSREFS Cf. A298678, A298680, A298681, A298682, A298683. Sequence in context: A297444 A184508 A101142 * A261885 A186750 A203715 Adjacent sequences:  A298676 A298677 A298678 * A298680 A298681 A298682 KEYWORD nonn,easy AUTHOR Felix Fröhlich, Jan 24 2018 EXTENSIONS More terms from Colin Barker, Jan 25 2018 STATUS approved

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Last modified September 26 15:09 EDT 2022. Contains 357000 sequences. (Running on oeis4.)