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A049666
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a(n) = Fibonacci(5*n)/5.
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32
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0, 1, 11, 122, 1353, 15005, 166408, 1845493, 20466831, 226980634, 2517253805, 27916772489, 309601751184, 3433536035513, 38078498141827, 422297015595610, 4683345669693537, 51939099382224517, 576013438874163224
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OFFSET
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0,3
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COMMENTS
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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
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)
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LINKS
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FORMULA
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G.f.: x/(1 - 11*x - x^2).
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
Lim_{k->infinity} a(n+k)/a(k) = (A001946(n) + A049666(n)*sqrt(125))/2.
(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
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
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EXAMPLE
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G.f. = x + 11*x^2 + 122*x^3 + 1353*x^4 + 15005*x^5 + 166408*x^6 + ...
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MATHEMATICA
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Table[Fibonacci[5*n]/5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)
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PROG
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(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)
(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
(Magma) [Fibonacci(5*n)/5: n in [0..30]]; // G. C. Greubel, Dec 02 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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