A099279 Squares of A001076.

0, 1, 16, 289, 5184, 93025, 1669264, 29953729, 537497856, 9645007681, 173072640400, 3105662519521, 55728852710976, 1000013686278049, 17944517500293904, 322001301319012225, 5778078906241926144, 103683419011035658369, 1860523463292399924496, 33385738920252162982561

Offset

0,3

Comments

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.

Unsigned member r=-16 of the family of Chebyshev sequences S_r(n) defined in A092184.

(-1)^(n+1)*a(n) = S_{-16}(n), n >= 0, defined in A092184.

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

Links

G. C. Greubel, Table of n, a(n) for n = 0..750

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (17,17,-1).

Formula

a(n) = A001076(n)^2.

a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=16.

a(n) = 18*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2, a(0)=0, a(1)=1.

a(n) = (T(n, 9) - (-1)^n)/10 with Chebyshev's T(n, x) polynomials of the first kind. T(n, 9) = A023039(n).

G.f.: x*(1-x)/((1+x)*(1-18*x+x^2)) = x*(1-x)/(1-17*x-17*x^2+x^3).

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

a(n+1) = (1 + (-1)^n)/2 + 16*Sum_{r=1..n} ( r*a(n+1-r) ). - Michael A. Allen, Mar 12 2023

E.g.f.: exp(-x)*(exp(10*x)*cosh(4*sqrt(5)*x) - 1)/10. - Stefano Spezia, Apr 06 2023

Maple

with (combinat):seq(fibonacci(n, 4)^2, n=0..16); # Zerinvary Lajos, Apr 09 2008
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

Mathematica

LinearRecurrence[{17, 17, -1}, {0, 1, 16}, 30] (* Harvey P. Dale, Mar 26 2012 *)
Fibonacci[3*Range[0, 30]]^2/4 (* G. C. Greubel, Aug 18 2022 *)

Prog

(MuPAD) numlib::fibonacci(3*n)^2/4 $ n = 0..35; // Zerinvary Lajos, May 13 2008
(Sage) [(fibonacci(3*n))^2/4 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009
(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
(Magma) [Fibonacci(3*n)^2/4: n in [0..30]]; // G. C. Greubel, Aug 18 2022

Crossrefs

Cf. A000045, A001076, A001654, A023039, A053120, A092184.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, this sequence, A099365, A099366, A099367, A099369, A099372, A099374.

Sequence in context: A225194 A027776 A140770 * A202878 A372172 A183886

Adjacent sequences: A099276 A099277 A099278 * A099280 A099281 A099282

Keyword

nonn,easy

Author

Wolfdieter Lang, Oct 18 2004

Status

approved