OFFSET
0,1
COMMENTS
Apart from initial term, same as A001060.
Pisano period lengths same as for A001060. - R. J. Mathar, Aug 10 2012
The beginning of this sequence is the only sequence of four consecutive primes in a Fibonacci-type sequence. - Franklin T. Adams-Watters, Mar 21 2015
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A104449. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 2) by applying positive powers of the automorphic matrix A^k = Matrix([A(k), B(k)], [B(k), A(k+1)]), with A(k) = S(k-1, 3) - S(k-2, 3) and B(k) = S(k-1, 3). See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019
For n>1, a(n) is the number of ways to tile a strip of length n+1 with black and white squares and white dominos, where there must be exactly one black square and it must appear amongst the first three cells. - Greg Dresden and Emma Li, Aug 24 2024
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..2500
Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, and John Pryce, On Higher Dimensional Interlacing Fibonacci Sequences, Continued Fractions and Chebyshev Polynomials, arXiv:1502.03085 [math.NT], 2015 (see p. 31).
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, The Geometry of some Fibonacci Identities in the Hosoya Triangle, arXiv:1804.02481 [math.NT], 2018.
Tanya Khovanova, Recursive Sequences
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).
FORMULA
a(n) = a(n-1) + a(n-2).
a(n) = F(n+3) - F(n-2) for n>1, where F=A000045. - Gerald McGarvey, Jul 10 2004
a(n) = 2*F(n-3) + F(n) for n>1. - Zerinvary Lajos, Oct 05 2007
G.f.: (3-x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = Sum_{k = n-3..n+1} F(k). - Gary Detlefs, Dec 30 2012
a(n) = ((3*sqrt(5)+1)*(((1+sqrt(5))/2)^n)+(3*sqrt(5)-1)*(((1-sqrt(5))/2)^n))/(2*sqrt(5)). - Bogart B. Strauss, Jul 19 2013
a(n) = F(n+3) + F(n-3) - 3*F(n) for n>1. - Bruno Berselli, Dec 29 2016
Bisection: a(2*k) = 3*S(k, 3) - 4*S(k-1, 3), a(2*k+1) = 2*S(k, 3) + S(k-1, 3), for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= -1. - Wolfdieter Lang, May 28 2019
a(3n + 2)/a(3n - 1) = continued fraction 4,4,4,...,4,-5 (that's n 4's followed by a single -5). - Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: ((- 1 + 3*sqrt(5))*exp((1/2)*(1 - sqrt(5))*x) + (1 + 3*sqrt(5))*exp((1/2)*(1 + sqrt(5))*x))/(2*sqrt(5)). - Stefano Spezia, Jul 17 2019
a(n) = (F(3n+1) - F(n+1)^3)/(F(n)^2) for n>1, where F(n) = A000045(n). - Michael Tulskikh, Jul 22 2020
a(n) = 3 * Sum_{k=0..n-2} A168561(n-2,k) + 2 * Sum_{k=0..n-1} A168561(n-1,k), n>0. - R. J. Mathar, Feb 14 2024
MAPLE
with(combinat): a:=n->2*fibonacci(n-1)+fibonacci(n+2): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007
MATHEMATICA
LinearRecurrence[{1, 1}, {3, 2}, 40] (* or *)
Table[Fibonacci[n + 1] + LucasL[n], {n, 0, 40}] (* or *)
Table[Fibonacci[n + 3] + Fibonacci[n - 3] - 3*Fibonacci[n], {n, 2, 40}] (* Bruno Berselli, Dec 30 2016 *)
PROG
(Magma) [2*Fibonacci(n-3)+Fibonacci(n): n in [2..41]]; // Vincenzo Librandi, Apr 16 2011
(Magma) [GeneralizedFibonacciNumber(3, 2, n): n in [0..39]]; // Arkadiusz Wesolowski, Mar 16 2016
(PARI) a(n)=([0, 1; 1, 1]^n*[3; 2])[1, 1] \\ Charles R Greathouse IV, Sep 24 2015
(PARI) a(n)=2*fibonacci(n-3) + fibonacci(n) \\ Charles R Greathouse IV, Sep 24 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Definition corrected by Gary Detlefs, Dec 30 2012
STATUS
approved