cp's OEIS Frontend

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.

%I A298678 #22 Jan 29 2018 05:49:01
%S A298678 1,0,7,12,73,216,919,3204,12409,45408,171271,635580,2379241,8865000,
%T A298678 33113527,123523572,461111833,1720661616,6422058919,23966525484,
%U A298678 89446140169,333813840888,1245817611991,4649439829860,17351975261881,64758394108800,241681735391047
%N 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.
%C A298678 The following substitution rules apply to the tiles:
%C A298678 triangle with 6 markings -> 1 hexagon
%C A298678 triangle with 4 markings -> 1 square, 2 triangles with 4 markings
%C A298678 square                   -> 1 square, 4 triangles with 6 markings
%C A298678 hexagon                  -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
%C A298678 For n > 0, a(n) is also the number of triangles with 6 markings after n iterations when starting with the hexagon.
%C A298678 a(n) is also the number of triangles with 6 markings after n iterations when starting with the triangle with 6 markings.
%C A298678 a(n) is also the number of hexagons after n iterations when starting with the triangle with 6 markings.
%H A298678 Colin Barker, <a href="/A298678/b298678.txt">Table of n, a(n) for n = 0..1000</a>
%H A298678 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 A298678 Tilings Encyclopedia, <a href="https://tilings.math.uni-bielefeld.de/substitution/shield">Shield</a>
%H A298678 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,7,-2).
%F A298678 G.f.: (1-2*x)/((1+2*x)*(1-4*x+x^2)). - _Joerg Arndt_, Jan 25 2018
%F A298678 13*a(n) = A077235(n) + 8*(-2)^n. - _Bruno Berselli_, Jan 25 2018
%F A298678 From _Colin Barker_, Jan 25 2018: (Start)
%F A298678 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))).
%F A298678 a(n) = 2*a(n-1) + 7*a(n-2) - 2*a(n-3) for n>2.
%F A298678 (End)
%o A298678 (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 A298678 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 A298678 terms(n) = my(v=[0, 0, 0, 1], i=0); while(1, print1(v[4], ", "); i++; if(i==n, break, v=substitute(v)))
%o A298678 (PARI) Vec((1-2*x)/((1+2*x)*(1-4*x+x^2)) + O(x^40)) \\ _Colin Barker_, Jan 25 2018
%Y A298678 Cf. A298679, A298680, A298681, A298682, A298683.
%K A298678 nonn,easy
%O A298678 0,3
%A A298678 _Felix Fröhlich_, Jan 24 2018
%E A298678 More terms from _Colin Barker_, Jan 25 2018