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a(n) = Fibonacci(6*n)/8.
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%I #119 Jun 06 2022 08:15:58

%S 0,1,18,323,5796,104005,1866294,33489287,600940872,10783446409,

%T 193501094490,3472236254411,62306751484908,1118049290473933,

%U 20062580477045886,360008399296352015,6460088606857290384,115921586524134874897,2080128468827570457762

%N a(n) = Fibonacci(6*n)/8.

%C For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - _John M. Campbell_, Jul 08 2011

%C For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - _Milan Janjic_, Jan 25 2015

%C 10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - _Wolfdieter Lang_, Feb 01 2016

%C Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - _Herbert Kociemba_, Jun 05 2022

%H Indranil Ghosh, <a href="/A049660/b049660.txt">Table of n, a(n) for n = 0..796</a> (terms 0..200 from Vincenzo Librandi)

%H Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.

%H R. Flórez, R. A. Higuita, and A. Mukherjee, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mukherjee/mukh2.html">Alternating Sums in the Hosoya Polynomial Triangle</a>, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

%F G.f.: x/(1 - 18*x + x^2).

%F a(n) = A134492(n)/8.

%F a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002

%F For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - _Gregory V. Richardson_, Oct 14 2002

%F a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.

%F a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).

%F a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).

%F a(n) = 18*a(n-1) - a(n-2). - _Gregory V. Richardson_, Oct 14 2002

%F a(n) = A001076(2n)/4.

%F a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - _Mohamed Bouhamida_, May 26 2007

%F a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - _Philippe Deléham_, Feb 10 2012

%F Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - _Peter Bala_, Dec 23 2012

%F Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - _Peter Bala_, Dec 23 2012

%F a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).

%F Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - _Peter Bala_, Nov 29 2013

%F From _Peter Bala_, Apr 02 2015: (Start)

%F Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.

%F 1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.

%F 1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).

%F 1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.

%F 1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.

%F 1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).

%F 1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).

%F 1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).

%F 1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).

%F 1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).

%F 1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).

%F (End)

%e a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017

%p with (combinat):seq(fibonacci(2*n,4)/4, n=0..16); # _Zerinvary Lajos_, Apr 20 2008

%t Fibonacci[6*Range[0,20]]/8 (* _Harvey P. Dale_, Nov 23 2011 *)

%t LinearRecurrence[{18,-1}, {0,1}, 30] (* _G. C. Greubel_, Dec 02 2017 *)

%t Table[ChebyshevU[-1 + n, 9], {n, 0, 18}] (* _Herbert Kociemba_, Jun 05 2022 *)

%o (MuPAD) numlib::fibonacci(6*n)/8 $ n = 0..25; // _Zerinvary Lajos_, May 09 2008

%o (Sage) [lucas_number1(n,18,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 25 2008

%o (Sage) [fibonacci(6*n)/8 for n in range(0, 17)] # _Zerinvary Lajos_, May 15 2009

%o (PARI) a(n)=fibonacci(6*n)/8 \\ _Charles R Greathouse IV_, Apr 17 2012

%o (Magma) [Fibonacci(6*n)/8: n in [0..30]]; // _G. C. Greubel_, Dec 02 2017

%Y Column m=6 of array A028412.

%Y Partial sums of A007805.

%Y Cf. A000045, A001076, A001077, A014445, A014448, A015448, A099843, A134492, A155179.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_

%E Chebyshev and other comments from _Wolfdieter Lang_, Nov 08 2002