A298680 - OEIS

%I #15 Jan 29 2018 05:51:19

%S 0,0,4,12,56,192,756,2748,10408,38544,144452,537900,2009880,7496160,

%T 27985684,104424732,389756936,1454515632,5428480356,20259056268,

%U 75608443768,282173320704,1053087635252,3930171627900,14667610061160,54740246247120,204293419666564

%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 6 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 the number of hexagonal tiles after n+1 iterations when starting with the triangle with 4 markings.

%H Colin Barker, <a href="/A298680/b298680.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.: 4*x^2 / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).

%F a(n) = (1/39)*(-26 + (-1)^n*2^(3+n) - (2-sqrt(3))^n*(-9+sqrt(3)) + (2+sqrt(3))^n*(9+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[1], ", "); i++; if(i==n, break, v=substitute(v)))

%o (PARI) concat(vector(2), Vec(4*x^2 / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ _Colin Barker_, Jan 25 2018

%Y Cf. A298678, A298679, A298681, A298682, A298683.

%K nonn,easy

%O 0,3

%A _Felix Fröhlich_, Jan 24 2018

%E More terms from _Colin Barker_, Jan 25 2018