

A187498


Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p={p_1, p_2, p_3, p_4} = {3,0,1,2}, n=3*r+p_i, and define a(3)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).


7



0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 3, 4, 4, 6, 5, 10, 10, 14, 15, 20, 20, 34, 35, 48, 55, 69, 75, 117, 124, 165, 199, 241, 274, 406, 440, 571, 714, 846, 988, 1417, 1560, 1988, 2548, 2977, 3536, 4965, 5525, 6953, 9061, 10490, 12597, 17443, 19551
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,9


COMMENTS

Theory: (Start)
1. Definitions. Let T_(9,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/9,(9j)*Pi/9) and Area(T_(9,j,0))=sin(j*Pi/9), j in {1,2,3,4}. Associated with T_(9,j,0) are its angle coefficients (j, 9j) in which one coefficient is even while the other is odd. A halftile is created by cutting T_(9,j,0) along a line extending between its two corners with even angle coefficient; let H_(9,j,0) denote this halftile. Similarly, a T_(9,j,r) tile is a linearly scaled version of T_(9,j,0) with sides of length Q^r and Area(T_(9,j,r))=Q^(2*r)*sin(j*Pi/9), r>=0 an integer, where Q is the positive, constant square root Q=sqrt(2*cos(Pi/9)); likewise let H_(9,j,r) denote the corresponding halftile. Often H_(9,i,r) (i in {1,2,3,4}) can be subdivided into an integral number of each equivalence class H_(9,j,0). But regardless of whether or not H_(9,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3,4, in which the entry m_(i,j) gives the quantity of H_(9,j,0) tiles that should be present in a subdivided H_(9,i,r) tile. The number Q^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_1)^r, where
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
2. The sequence. Let r>=0, and let D_r be the rth "block" defined by D_r={a(3*r3),a(3*r),a(3*r+1),a(3*r+2)} with a(3)=0. Note that D_rD_(r1)3*D_(r2)+2*D_(r3)+D_(r4)={0,0,0,0}, for r>=4, with initial conditions {D_k}={{0,0,0,1},{0,0,1,1},{0,1,1,2},{1,1,3,3}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={3,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block D_r corresponds componentwise to the fourth column of M, and a(3*r+p_i)=m_(i,4) gives the quantity of H_(9,4,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
Combining blocks A_r, B_r, C_r and D_r, from A187495, A187496, A187497 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_1)^r, and a negative index (1)*r yields the corresponding inverse [A_(r),B_(r),C_(r),D_(r)]=(U_1)^(r) of (U_1)^r.. Therefore the four sequences need not be causal.
Since U_1 is symmetric, so is M=(U_1)^r, so the block D_r also corresponds to the fourth row of M. Therefore, alternatively, for j=1,2,3,4, a(3r+p_j)=m_(4,j) gives the quantity of H_(9,j,0) tiles that should be present in a H_(9,4,r) tile.
Since a(3*r)=a(3*(r+1)3) for all r, this sequence arises by concatenation of fourthcolumn entries m_(2,4), m_(3,4) and m_(4,4) (or fourthrow entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_1)^r.


REFERENCES

L. E. Jeffery, Unitprimitive matrices and rhombus substitution tilings, (in preparation).


LINKS



FORMULA

Recurrence: a(n) = a(n3) +3*a(n6) 2*a(n9) a(n12), for n >= 12, with initial conditions {a(m)} = {0,0,1,0,1,1,1,1,2,1,3,3}, m=0,1,...,11.
G.f.: x^2*(1+x)*(x^6+3*x^4+2*x^2+1) / ( (1+x+x^2)*(x^9+3*x^61) ).


MAPLE

A052931 := proc(n) if n < 0 then 0; else coeftayl(1/(13*x^2x^3), x=0, n) ; end if; end proc:
A052931a := proc(n) if n mod 3 = 0 then A052931(n/3) ; else 0 ; end if; end proc:
A057078 := proc(n) op(1+(n mod 3), [1, 0, 1]) ; end proc:
A187498 := proc(n) A057078(n) +A052931a(n) +2*A052931a(n2) +A052931a(n3) +3*A052931a(n4) +2*A052931a(n5) +A052931a(n6) +3*A052931a(n7) A052931a(n8) ; %/3 ; end proc:


MATHEMATICA

CoefficientList[Series[x^2*(1 + x)*(x^6 + 3*x^4 + 2*x^2 + 1)/((1 + x + x^2)*(x^9 + 3*x^6  1)), {x, 0, 1000}], x] (* G. C. Greubel, Sep 23 2017 *)


PROG

(PARI) x='x+O('x^50); Vec(x^2*(1+x)*(x^6+3*x^4+2*x^2+1)/((1+x+x^2)*(x^9+3*x^61))) \\ G. C. Greubel, Sep 23 2017


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



