cp's OEIS Frontend

A298683 Start with the square tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of squares after n iterations.

%I A298683 #27 Jun 27 2025 22:59:12
%S A298683 1,1,1,13,37,169,577,2269,8245,31225,115633,433357,1613701,6029641,
%T A298683 22488481,83957053,313274197,1169270809,4363546897,16285441069,
%U A298683 60777168805,226825331305,846519962113,3159262905757,11790514883701,44002830183481,164220738741361
%N A298683 Start with the square tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of squares after n iterations.
%C A298683 The following substitution rules apply to the tiles:
%C A298683 triangle with 6 markings -> 1 hexagon
%C A298683 triangle with 4 markings -> 1 square, 2 triangles with 4 markings
%C A298683 square                   -> 1 square, 4 triangles with 6 markings
%C A298683 hexagon                  -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
%C A298683 a(n) is also one more than the number of triangles with 4 markings after n iterations when starting with the square tile.
%H A298683 Colin Barker, <a href="/A298683/b298683.txt">Table of n, a(n) for n = 0..1000</a>
%H A298683 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 A298683 Tilings Encyclopedia, <a href="https://tilings.math.uni-bielefeld.de/substitution/shield">Shield</a>
%H A298683 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,5,-9,2).
%F A298683 G.f.: ((1-2*x)*(1-7*x^2))/((1-x)*(1+2*x)*(1-4*x+x^2)). - _Joerg Arndt_, Jan 25 2018
%F A298683 From _Colin Barker_, Jan 25 2018: (Start)
%F A298683 a(n) = (1/13)*(-13 + (-1)^(1+n)*2^(2+n) + (15-7*sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(15+7*sqrt(3))).
%F A298683 a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
%F A298683 (End)
%F A298683 a(n) = ((15 - 7*sqrt(3))*(2 + sqrt(3))^n + (2 - sqrt(3))^n*(15 + 7*sqrt(3)) - 4*(-2)^n)/13 - 1. - _Bruno Berselli_, Jan 25 2018
%t A298683 CoefficientList[Series[((1 - 2 x) (1 - 7 x^2))/((1 - x) (1 + 2 x) (1 - 4 x + x^2)), {x, 0, 26}], x] (* or *)
%t A298683 LinearRecurrence[{3, 5, -9, 2}, {1, 1, 1, 13}, 27] (* _Michael De Vlieger_, Jan 28 2018 *)
%t A298683 f[n_] := Simplify[(-13 + (-1)^(n + 1)*2^(2 + n) + (15 - 7 Sqrt[3])*(2 + Sqrt[3])^n + (2 - Sqrt[3])^n*(15 + 7 Sqrt[3]))/13]; Array[f, 28, 0] (* _Robert G. Wilson v_, Feb 26 2018 *)
%o A298683 (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 A298683 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 A298683 terms(n) = my(v=[0, 0, 1, 0], i=0); while(1, print1(v[3], ", "); i++; if(i==n, break, v=substitute(v)))
%o A298683 (PARI) Vec(((1-2*x)*(1-7*x^2))/((1-x)*(1+2*x)*(1-4*x+x^2)) + O(x^40)) \\ _Colin Barker_, Jan 25 2018
%Y A298683 Cf. A298678, A298679, A298680, A298681, A298682.
%K A298683 nonn,easy
%O A298683 0,4
%A A298683 _Felix Fröhlich_, Jan 24 2018