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%I #59 Feb 23 2024 07:30:17
%S 0,1,16,289,5184,93025,1669264,29953729,537497856,9645007681,
%T 173072640400,3105662519521,55728852710976,1000013686278049,
%U 17944517500293904,322001301319012225,5778078906241926144,103683419011035658369,1860523463292399924496,33385738920252162982561
%N Squares of A001076.
%C For the generalized Fibonacci sequences U(n-1;a) = (ap(a)^n - am(a)^n)/(ap(a) - am(a)) with ap(a) = (a + sqrt(a^2+4))/2, am(a) = (a - sqrt(a^2+4))/2, a from the integers, one has for the squared sequences U(n-1;a)^2 = (2*T(n,(a^2+2)/2) - 2*(-1)^n)/(a^2+4). Here T(n,x) are Chebyshev's polynomials of the first kind (see A053120). Therefore the o.g.f. for the squared sequence is x*(1-x)/((1+x)*(1-(a^2+2)*x+x^2)) = x*(1-x)/(1 - (a^2+1)*x - (a^2+1)*x^2 + x^3). For this example a=4.
%C Unsigned member r=-16 of the family of Chebyshev sequences S_r(n) defined in A092184.
%C (-1)^(n+1)*a(n) = S_{-16}(n), n >= 0, defined in A092184.
%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 4 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 4 kinds of (1/4,1/4)-fences available. - _Michael A. Allen_, Mar 12 2023
%H G. C. Greubel, <a href="/A099279/b099279.txt">Table of n, a(n) for n = 0..750</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/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (17,17,-1).
%F a(n) = A001076(n)^2.
%F a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=16.
%F a(n) = 18*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2, a(0)=0, a(1)=1.
%F a(n) = (T(n, 9) - (-1)^n)/10 with Chebyshev's T(n, x) polynomials of the first kind. T(n, 9) = A023039(n).
%F G.f.: x*(1-x)/((1+x)*(1-18*x+x^2)) = x*(1-x)/(1-17*x-17*x^2+x^3).
%F a(n) = a(n-1) + A001654(3*n-2) with a(0)=0, where A001654 are the golden rectangle numbers. - _Johannes W. Meijer_, Sep 22 2010
%F a(n+1) = (1 + (-1)^n)/2 + 16*Sum_{r=1..n} ( r*a(n+1-r) ). - _Michael A. Allen_, Mar 12 2023
%F E.g.f.: exp(-x)*(exp(10*x)*cosh(4*sqrt(5)*x) - 1)/10. - _Stefano Spezia_, Apr 06 2023
%p with (combinat):seq(fibonacci(n,4)^2,n=0..16); # _Zerinvary Lajos_, Apr 09 2008
%p nmax:=48: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax/3 do a(n):=a(n-1)+A001654(3*n-2) od: seq(a(n),n=0..nmax/3); # _Johannes W. Meijer_, Sep 22 2010
%t LinearRecurrence[{17,17,-1},{0,1,16},30] (* _Harvey P. Dale_, Mar 26 2012 *)
%t Fibonacci[3*Range[0, 30]]^2/4 (* _G. C. Greubel_, Aug 18 2022 *)
%o (MuPAD) numlib::fibonacci(3*n)^2/4 $ n = 0..35; // _Zerinvary Lajos_, May 13 2008
%o (Sage) [(fibonacci(3*n))^2/4 for n in range(0, 17)] # _Zerinvary Lajos_, May 15 2009
%o (PARI) my(x='x+O('x^99)); concat([0], Vec(x*(1-x)/((1-18*x+x^2)*(1+x)))) \\ _Altug Alkan_, Dec 17 2017
%o (Magma) [Fibonacci(3*n)^2/4: n in [0..30]]; // _G. C. Greubel_, Aug 18 2022
%Y Cf. A000045, A001076, A001654, A023039, A053120, A092184.
%Y Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, this sequence, A099365, A099366, A099367, A099369, A099372, A099374.
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Oct 18 2004
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