A049666 a(n) = Fibonacci(5*n)/5.

0, 1, 11, 122, 1353, 15005, 166408, 1845493, 20466831, 226980634, 2517253805, 27916772489, 309601751184, 3433536035513, 38078498141827, 422297015595610, 4683345669693537, 51939099382224517, 576013438874163224

Offset

0,3

Comments

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 11's along the main diagonal and 1's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 08 2011

For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,11} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

For n >= 1, a(n) equals the denominator of the continued fraction [11, 11, ..., 11] (with n copies of 11). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019

From Michael A. Allen, Mar 30 2023: (Start)

Also called the 11-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 11 kinds of squares available. (End)

Links

G. C. Greubel, Table of n, a(n) for n = 0..950

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Tanya Khovanova, Recursive Sequences

Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.

Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.

Index entries for linear recurrences with constant coefficients, signature (11,1).

Formula

G.f.: x/(1 - 11*x - x^2).

a(n) = A102312(n)/5.

a(n) = 11*a(n-1) + a(n-2) for n > 1, a(0)=0, a(1)=1. With a=golden ratio and b=1-a, a(n) = (a^(5n)-b^(5n))/(5*sqrt(5)). - Mario Catalani (mario.catalani(AT)unito.it), Jul 24 2003

a(n) = F(n, 11), the n-th Fibonacci polynomial evaluated at x=11. - T. D. Noe, Jan 19 2006

a(n) = ((11+sqrt(125))^n-(11-sqrt(125))^n)/(2^n*sqrt(125)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009

From Johannes W. Meijer, Jun 12 2010: (Start)

a(2n) = 11*A049670(n), a(2n+1) = A097843(n).

a(3n+1) = A041227(5n), a(3n+2) = A041227(5n+3), a(3n+3) = 2*A041227(5n+4).

Lim_{k->infinity} a(n+k)/a(k) = (A001946(n) + A049666(n)*sqrt(125))/2.

Lim_{n->infinity} A001946(n)/A049666(n) = sqrt(125).

(End)

a(n) = F(n) + (-1)^n*5*F(n)^3 + 5*F(n)^5, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012

a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014

E.g.f.: (exp((1/2)*(11-5*sqrt(5))*x)*(-1 + exp(5*sqrt(5)*x)))/(5*sqrt(5)). - Stefano Spezia, Aug 02 2019

Example

G.f. = x + 11*x^2 + 122*x^3 + 1353*x^4 + 15005*x^5 + 166408*x^6 + ...

Maple

A049666 := proc(n)
    combinat[fibonacci](5*n)/5 ;
end proc: # R. J. Mathar, May 07 2024

Mathematica

Table[Fibonacci[5*n]/5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)
a[ n_] := Fibonacci[n, 11]; (* Michael Somos, May 28 2014 *)

Prog

(MuPAD) numlib::fibonacci(5*n)/5 $ n = 0..25; // Zerinvary Lajos, May 09 2008
(Sage)
from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0, 1, 11, 11, 1, 0)
[next(it) for i in range(1, 22)] # Zerinvary Lajos, Jul 09 2008
(Sage) [lucas_number1(n, 11, -1) for n in range(0, 19)] # Zerinvary Lajos, Apr 27 2009
(Sage) [fibonacci(5*n)/5 for n in range(0, 19)] # Zerinvary Lajos, May 15 2009
(PARI) a(n)=fibonacci(5*n)/5 \\ Charles R Greathouse IV, Feb 03 2014
(Magma) [Fibonacci(5*n)/5: n in [0..30]]; // G. C. Greubel, Dec 02 2017

Crossrefs

A column of array A028412.

Cf. A000045, A102312, A243399.

Row n=11 of A073133, A172236 and A352361, and column k=11 of A157103.

Sequence in context: A293805 A358697 A288791 * A163462 A334000 A041222

Adjacent sequences: A049663 A049664 A049665 * A049667 A049668 A049669

Keyword

nonn,easy,changed

Author

Clark Kimberling

Status

approved