A187069 Let i be in {1,2,3}, let r >= 0 be an integer and n=2*r+i-1. Then a(n)=a(2*r+i-1) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

0, 1, 0, 1, 1, 2, 2, 4, 5, 9, 11, 20, 25, 45, 56, 101, 126, 227, 283, 510, 636, 1146, 1429, 2575, 3211, 5786, 7215, 13001, 16212, 29213, 36428, 65641, 81853, 147494, 183922, 331416, 413269, 744685, 928607, 1673292, 2086561, 3759853, 4688460, 8448313, 10534874

Offset: 0

Author

L. Edson Jeffery, Mar 06 2011

Keywords

• nonn
• easy

Comments

See A187070 for supporting theory. Define the matrix

U_2 = (0 0 1)

(0 1 1)

(1 1 1).

Let r>=0, and let B_r be the r-th "block" defined by B_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that B_r-2*B_(r-1)-B_(r-2)+B_(r-3)={0,0,0}. Let n=2*r+i-1 and M=(m_(i,j))=(U_2)^r. Then B_r corresponds component-wise to the second column of M, and a(n)=a(2*r+i-1)=m_(i,2) gives the quantity of H_(7,2,0) tiles that should appear in a subdivided H_(7,i,r) tile.

Since a(2*r+2)=a(2*(r+1)) for all r, this sequence arises by concatenation of second-column entries m_(1,2) and m_(2,2) from successive matrices M=(U_2)^r.

Examples

			Suppose r=3.
Then B_r = B_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {2,4,5}, corresponding to the entries in the second column of
  M = (U_2)^3 = (1 2 3)
                (2 4 5)
                (3 5 6).
Suppose i=2. Setting n=2*r+i-1, then a(n) = a(2*r+i-1) = a(6+2-1) = a(7) = m_(2,2) = 4. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,2) = 4 H_(7,2,0) tiles.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• L. Edson Jeffery, Unit-primitive matrices
• Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
• Index entries for linear recurrences with constant coefficients, signature (0,2,0,1,0,-1).

Crossrefs

Cf. A038196, A187068, A187070, A006054.

Programs

Mathematica

CoefficientList[Series[x*(1 - x^2 + x^3 - x^4)/(1 - 2*x^2 - x^4 + x^6), {x, 0, 50}], x] (* G. C. Greubel, Oct 20 2017 *)
LinearRecurrence[{0,2,0,1,0,-1},{0,1,0,1,1,2},50] (* Harvey P. Dale, Dec 16 2017 *)

PARI

my(x='x+O('x^50)); concat([0], Vec(x*(1-x^2+x^3-x^4)/(1-2*x^2-x^4+x^6))) \\ G. C. Greubel, Oct 20 2017

Formula

Recurrence: a(n) = 2*a(n-2) + a(n-4) - a(n-6).

G.f.: x*(1-x^2+x^3-x^4)/(1-2*x^2-x^4+x^6).

Closed-form: a(n) = (1/14)*[[X_1+Y_1*(-1)^(n-1)]*[(w_2)^2-(w_3)^2]*(w_1)^(n-1)+[X_2+Y_2*(-1)^(n-1)]*[(w_3)^2-(w_1)^2]*(w_2)^(n-1)+[X_3+Y_3*(-1)^(n-1)]*[(w_1)^2-(w_2)^2]*(w_3)^(n-1)], where w_k = sqrt[(2cos(k*Pi/7))^2-1], X_k = (w_k)^4-(w_k)^2+w_k-1 and Y_k = (w_k)^4+(w_k)^2-w_k-1, k=1,2,3.

a(2*n) = A006054(n), a(2*n+3) = A052534(n).