A092936 Area of n-th triple of hexagons around a triangle.

1, 9, 100, 1089, 11881, 129600, 1413721, 15421329, 168220900, 1835008569, 20016873361, 218350598400, 2381839709041, 25981886201049, 283418908502500, 3091626107326449, 33724468272088441, 367877524885646400

Offset: 1

Author

Peter J. C. Moses, Apr 18 2004

Keywords

• easy
• nonn

Comments

This is the unsigned member r=-9 of the family of Chebyshev sequences S_r(n) defined in A092184: ((-1)^(n+1))*a(n) = S_{-9}(n), n>=0.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, red half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), green half-squares, and blue half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,3/4)-fences, red (1/4,1/4)-fences, green (1/4,1/4)-fences, and blue (1/4,1/4)-fences. - Michael A. Allen, Dec 30 2022

Examples

			a(5) = 10*(1089+100)-9 = 11881. From A006190, a(5) = (3*33+10)^2 = 11881.
		

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..200
• Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
• Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
• Index entries for linear recurrences with constant coefficients, signature (10,10,-1).

Crossrefs

Cf. A005386, A006190, A057076, A092184, A176019.

Programs

GAP

a:=[1,9,100];; for n in [4..18] do a[n]:=10*(a[n-1]+a[n-2])-a[n-3]; od; a; # Muniru A Asiru, Feb 20 2018

Maple

seq(fibonacci(n,3)^2,n=1..18); # Zerinvary Lajos, Apr 05 2008

Mathematica

CoefficientList[Series[(1-x)*x/(1-10*x-10*x^2+x^3), {x, 0, 20}], x]
(CoefficientList[Series[x/(1-3*x-x^2), {x, 0, 20}], x])^2
Table[Round[((3+Sqrt[13])^n)^2/(13*4^n)], {n, 0, 20}]
LinearRecurrence[{10, 10, -1}, {1, 9, 100}, 18] (* Georg Fischer, Feb 22 2019 *)

Formula

a(n) = A006190(n)^2.

a(n) = 10*(a(n-1)+a(n-2)) - a(n-3).

G.f.: (1-x)*x/(1-10*x-10*x^2+x^3).

a(n) = ((3-sqrt(13))^n-(3+sqrt(13))^n)^2/(13*4^n).

a(n) = 2*(T(n, 11/2)-(-1)^n)/13 with twice the Chebyshev polynomials of the first kind evaluated at x = 11/2: 2*T(n, 11/2) = A057076(n) = ((11+3*sqrt(13))^n + (11-3*sqrt(13))^n)/2^n. - Wolfdieter Lang, Oct 18 2004

From Michael A. Allen, Dec 30 2022: (Start)

a(n+1) = 11*a(n) - a(n-1) + 2*(-1)^n.

a(n+1) = (1 + (-1)^n)/2 + 9*Sum_{k=1..n} ( k*a(n+1-k) ). (End)

Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(13))/6 (A176019) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024