

A033890


a(n) = Fibonacci(4*n+2).


37



1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

a(n) = S(n,7)+S(n1,7) = S(2*n,sqrt(9) = 3), 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).
(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
This sequence consists of the oddindexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The evenindexed terms of A001906 are in A033888. Lim. n> Inf. a(n)/a(n1) = phi^4 = (7 + 3*Sqrt(5))/2.  Gregory V. Richardson, Oct 13 2002
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
General recurrence is a(n)=(a(1)1)*a(n1)a(n2), 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
Indices of square numbers which are also 12gonal.  Sture Sjöstedt, Jun 01 2009
a(n) = A167816(4*n+2).  Reinhard Zumkeller, Nov 13 2009
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
If we let b(0) = 0 and, for n >=1, b(n) = A033890(n1), then the sequence b(n) will be F(4n2) and the first difference is L(4n) or A056854. F(4n2) 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
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourthorder 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 3parameter family of divisibility sequences found by Williams and Guy. See A100047.  Peter Bala, Mar 22 2015


REFERENCES

Fink, Alex, Richard Guy, and Mark Krusemeyer. "Partitions with parts occurring at most thrice." Contributions to Discrete Mathematics 3.2 (2008), 76114. See Section 13.


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Nathan D. Cahill and Darren A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216221. See p. 219.
Tanya Khovanova, Recursive Sequences
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160174.
Kival Ngaokrajang, Illustration of golden spiral length and pitch ratio
H. C. Williams and R. K. Guy, Some fourthorder linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 12551277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (7,1).
Index entries for sequences related to Chebyshev polynomials.


FORMULA

G.f.: (1+x)/(17*x+x^2).
a(n) = 7*a(n1)a(n2), n>1. a(0)=1, a(1)=8.
a(n) = [ [(7+3*Sqrt(5))^n  [(73*Sqrt(5))^n] + 2*[(7+3*Sqrt(5))^(n1)  [(73*Sqrt(5))^(n1)] ] / (3*(2^n)*Sqrt(5)).  Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0.. n} x^(ni)*binomial(2*ni, i); then (1)^n*q(n, 9) = a(n).  Benoit Cloitre, Nov 10 2002
Define f[x,s] = s x + Sqrt[(s^21)x^2+1]; f[0,s]=0. a(n) = f[a(n1),7/2] + f[a(n2),7/2].  Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n)8*a(n1)+ a(n2), a(1)=1, a(2)=8, a(3)=55.  Sture Sjöstedt, May 27 2009
a(n)=b such that (1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2sin(x)) dx = c + b*log(3).  Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)).  Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1).  R. J. Mathar, Apr 30 2017


MAPLE

A033890 := proc(n)
option remember;
if n <= 1 then
op(n+1, [1, 8]);
else
7*procname(n1)procname(n2) ;
end if;
end proc: # R. J. Mathar, Apr 30 2017


MATHEMATICA

Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7, 1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)


PROG

(PARI) a(n)=fibonacci(4*n+2)
(Sage) [(lucas_number2(n, 7, 1)lucas_number2(n1, 7, 1))/5 for n in xrange(1, 21)] # Zerinvary Lajos, Nov 10 2009
(MAGMA) [Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011


CROSSREFS

Cf. A000045, A001906, A100047.
Sequence in context: A154245 A143420 A075734 * A010924 A010918 A019484
Adjacent sequences: A033887 A033888 A033889 * A033891 A033892 A033893


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



