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A013655 a(n) = F(n+1) + L(n), where F(n) and L(n) are Fibonacci and Lucas numbers, respectively. 28
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 (list; graph; refs; listen; history; text; internal format)
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
LINKS
Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, 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, 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.
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
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
Sequence in context: A171018 A350192 A239260 * A223701 A220519 A094894
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Definition corrected by Gary Detlefs, Dec 30 2012
STATUS
approved

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Last modified June 24 11:11 EDT 2024. Contains 373677 sequences. (Running on oeis4.)