A187068 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_(7,1,0) 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).

1, 0, 0, 0, 1, 1, 1, 2, 3, 5, 6, 11, 14, 25, 31, 56, 70, 126, 157, 283, 353, 636, 793, 1429, 1782, 3211, 4004, 7215, 8997, 16212, 20216, 36428, 45425, 81853, 102069, 183922, 229347, 413269, 515338, 928607, 1157954, 2086561, 2601899

Offset: 0

Author

L. Edson Jeffery, Mar 06 2011

Keywords

• nonn
• easy

Comments

(Start) See A187070 for supporting theory. Define the matrix

U_2=

(0 0 1)

(0 1 1)

(1 1 1).

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

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

This sequence is a nontrivial extension of both A038196 and A187070.

Examples

			(Start) Suppose r=3. Then
A_r = A_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {1,2,3},
corresponding to the entries in the first column of
M = m_(i,j) = (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_(2,1) = 2. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,1) = 2 H_(7,1,0) tiles. (End)
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• L. E. 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, A187069, A187070.

Programs

Mathematica

a[0] = 1; a[1] = a[2] = a[3] = 0; a[4] = a[5] = 1; a[?Negative] = 0; a[n] := a[n] = 2*a[n-2] + a[n-4] - a[n-6]; Table[a[n], {n, 0, 42}] (* Jean-François Alcover, Jan 02 2013 *)
CoefficientList[Series[(1 - 2*x^2 + x^5)/(1 - 2*x^2 - x^4 + x^6), {x, 0, 50}], x] (* G. C. Greubel, Jul 06 2017 *)

PARI

x='x+O('x^50); Vec((1-2*x^2+x^5)/(1-2*x^2-x^4+x^6)) \\ G. C. Greubel, Jul 06 2017

Formula

{a(n+2)} = A187070.

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

G.f.: (1-2*x^2+x^5)/(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)^5-2*(w_k)^3+1 and Y_k = -(w_k)^5+2*(w_k)^3+1, k=1,2,3.