%I #113 May 13 2024 02:12:26
%S 1,8,55,377,2584,17711,121393,832040,5702887,39088169,267914296,
%T 1836311903,12586269025,86267571272,591286729879,4052739537881,
%U 27777890035288,190392490709135,1304969544928657,8944394323791464,61305790721611591,420196140727489673
%N a(n) = Fibonacci(4*n + 2).
%C (x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - _Floor van Lamoen_, Dec 10 2001
%C This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*Sqrt(5))/2. - _Gregory V. Richardson_, Oct 13 2002
%C General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - _Ctibor O. Zizka_, Sep 02 2008
%C Indices of square numbers which are also 12-gonal. - _Sture Sjöstedt_, Jun 01 2009
%C For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3'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 If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - _Kival Ngaokrajang_, Nov 03 2013
%C The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - _Peter Bala_, Mar 22 2015
%C Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - _Bernard Schott_, Mar 19 2021
%H G. C. Greubel, <a href="/A033890/b033890.txt">Table of n, a(n) for n = 0..1000</a>
%H Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, <a href="https://www.emis.de/journals/INTEGERS/papers/p38/p38.Abstract.html">Polynomial sequences on quadratic curves</a>, Integers, Vol. 15, 2015, #A38.
%H Nathan D. Cahill and Darren A. Narayan, <a href="http://www.fq.math.ca/Papers1/42-3/quartcahill03_2004.pdf">Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants</a>, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216-221. See p. 219.
%H Alex Fink, Richard K. Guy, and Mark Krusemeyer, <a href="https://doi.org/10.11575/cdm.v3i2.61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Donatella Merlini and Renzo Sprugnoli, <a href="https://doi.org/10.1016/j.disc.2016.08.017">Arithmetic into geometric progressions through Riordan arrays</a>, Discrete Mathematics 340.2 (2017): 160-174.
%H Kival Ngaokrajang, <a href="/A033890/a033890.pdf">Illustration of golden spiral length and pitch ratio</a>
%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
%H H. C. Williams and R. K. Guy, <a href="http://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a>, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-1).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F G.f.: (1+x)/(1-7*x+x^2).
%F a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
%F a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
%F a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - _Gregory V. Richardson_, Oct 13 2002
%F Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - _Benoit Cloitre_, Nov 10 2002
%F a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - _Reinhard Zumkeller_, Jun 01 2005
%F Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - _Marcos Carreira_, Dec 27 2006
%F a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - _Sture Sjöstedt_, May 27 2009
%F a(n) = A167816(4*n+2). - _Reinhard Zumkeller_, Nov 13 2009
%F a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - _Francesco Daddi_, Aug 01 2011
%F a(n) = A000045(A016825(n)). - _Michel Marcus_, Mar 22 2015
%F a(n) = A001906(2*n+1). - _R. J. Mathar_, Apr 30 2017
%p A033890 := proc(n)
%p option remember;
%p if n <= 1 then
%p op(n+1,[1,8]);
%p else
%p 7*procname(n-1)-procname(n-2) ;
%p end if;
%p end proc: # _R. J. Mathar_, Apr 30 2017
%t Table[Fibonacci[4n + 2], {n, 0, 14}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 21 2008 *)
%t LinearRecurrence[{7, -1}, {1, 8}, 50] (* _G. C. Greubel_, Jul 13 2017 *)
%o (PARI) a(n)=fibonacci(4*n+2);
%o (Magma) [Fibonacci(4*n +2): n in [0..100]]; // _Vincenzo Librandi_, Apr 17 2011
%Y Cf. A000045, A001906, A100047.
%Y Cf. A342709, A342710.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_