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a(n) = 2*(T(n, 11/2)-(-1)^n)/13 with twice the Chebyshev's 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
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<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10, 10, -1).
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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
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.
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)
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