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%I #28 Apr 13 2024 19:14:46
%S 0,1,100,10201,1040400,106110601,10822240900,1103762461201,
%T 112572948801600,11481337015302001,1170983802612002500,
%U 119428866529408953001,12180573402197101203600
%N a(n) = A041041(n-1)^2, n >= 1.
%C See the comment in A099279. This is example a=10.
%C a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 10 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 10 kinds of (1/4,1/4)-fences available. - _Michael A. Allen_, Mar 21 2024
%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 (101,101,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.
%F a(n) = A041041(n-1)^2, n >= 1, a(0)=0.
%F a(n) = 101*a(n-1) + 101*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=100.
%F a(n) = 102*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
%F a(n) = (T(n, 51) - (-1)^n)/52 with the Chebyshev polynomials of the first kind: T(n, 51) = (n).
%F G.f.: x*(1-x)/((1-102*x+x^2)*(1+x)) = x*(1-x)/(1-101*x-101*x^2+x^3).
%F a(n) = (1 - (-1)^n)/2 + 100*Sum_{r=1..n-1} r*a(n-r). - _Michael A. Allen_, Mar 21 2024
%t LinearRecurrence[{101,101,-1},{0,1,100},20] (* _Harvey P. Dale_, Nov 10 2021 *)
%Y Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, A099369, A099372, this sequence.
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 18 2004
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