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%I #37 Mar 21 2024 06:48:19
%S 1,9,100,1089,11881,129600,1413721,15421329,168220900,1835008569,
%T 20016873361,218350598400,2381839709041,25981886201049,
%U 283418908502500,3091626107326449,33724468272088441,367877524885646400
%N Area of n-th triple of hexagons around a triangle.
%C 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.
%C 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
%H Muniru A Asiru, <a href="/A092936/b092936.txt">Table of n, a(n) for n = 1..200</a>
%H Michael A. Allen and Kenneth Edwards, <a href="https://www.fq.math.ca/Papers1/60-5/allen.pdf">Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,10,-1).
%F a(n) = 10*(a(n-1)+a(n-2)) - a(n-3).
%F G.f.: (1-x)*x/(1-10*x-10*x^2+x^3).
%F a(n) = ((3-sqrt(13))^n-(3+sqrt(13))^n)^2/(13*4^n).
%F 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
%F From _Michael A. Allen_, Dec 30 2022: (Start)
%F a(n+1) = 11*a(n) - a(n-1) + 2*(-1)^n.
%F a(n+1) = (1 + (-1)^n)/2 + 9*Sum_{k=1..n} ( k*a(n+1-k) ). (End)
%e a(5) = 10*(1089+100)-9 = 11881. From A006190, a(5) = (3*33+10)^2 = 11881.
%p seq(fibonacci(n,3)^2,n=1..18); # _Zerinvary Lajos_, Apr 05 2008
%t CoefficientList[Series[(1-x)*x/(1-10*x-10*x^2+x^3), {x, 0, 20}], x]
%t (CoefficientList[Series[x/(1-3*x-x^2), {x, 0, 20}], x])^2
%t Table[Round[((3+Sqrt[13])^n)^2/(13*4^n)], {n, 0, 20}]
%t LinearRecurrence[{10, 10, -1}, {1, 9, 100}, 18] (* _Georg Fischer_, Feb 22 2019 *)
%o (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
%Y Equals (A006190)^2.
%Y Cf. A005386, A006190.
%K easy,nonn
%O 1,2
%A _Peter J. C. Moses_, Apr 18 2004
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