A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively.

3, 2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804, 6155, 9959, 16114, 26073, 42187, 68260, 110447, 178707, 289154, 467861, 757015, 1224876, 1981891, 3206767, 5188658, 8395425, 13584083, 21979508, 35563591, 57543099, 93106690, 150649789, 243756479

Offset: 0

Author

Mohammad K. Azarian

Keywords

• nonn
• easy

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).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Cf. A000045, A001060, A001906, A089270, A104449.

Programs

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

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 *)

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

Formula

a(n) = A000045(n+1) + A000032(n).

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

Extensions

Definition corrected by Gary Detlefs, Dec 30 2012