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A049666 a(n) = Fibonacci(5*n)/5. 32

%I

%S 0,1,11,122,1353,15005,166408,1845493,20466831,226980634,2517253805,

%T 27916772489,309601751184,3433536035513,38078498141827,

%U 422297015595610,4683345669693537,51939099382224517,576013438874163224

%N a(n) = Fibonacci(5*n)/5.

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

%C 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

%C 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

%C 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

%H G. C. Greubel, <a href="/A049666/b049666.txt">Table of n, a(n) for n = 0..950</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Shaoxiong Yuan, <a href="https://arxiv.org/abs/1907.12459">Generalized Identities of Certain Continued Fractions</a>, arXiv:1907.12459 [math.NT], 2019.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, 1).

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

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

%F 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

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

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

%F From _Johannes W. Meijer_, Jun 12 2010: (Start)

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

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

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

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

%F (End)

%F 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

%F a(-n) = -(-1)^n * a(n). - _Michael Somos_, May 28 2014

%F 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

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

%t Table[Fibonacci[5*n]/5, {n, 0, 100}] (* _T. D. Noe_, Oct 29 2009 *)

%t a[ n_] := Fibonacci[n, 11]; (* _Michael Somos_, May 28 2014 *)

%o (MuPAD) numlib::fibonacci(5*n)/5 $ n = 0..25; // _Zerinvary Lajos_, May 09 2008

%o (Sage) from sage.combinat.sloane_functions import recur_gen3; it = recur_gen3(0,1,11,11,1,0); [it.next() for i in xrange(1,22)] # _Zerinvary Lajos_, Jul 09 2008

%o (Sage) [lucas_number1(n,11,-1) for n in xrange(0, 19)] # _Zerinvary Lajos_, Apr 27 2009

%o (Sage) [fibonacci(5*n)/5 for n in xrange(0, 19)] # _Zerinvary Lajos_, May 15 2009

%o (PARI) a(n)=fibonacci(5*n)/5 \\ _Charles R Greathouse IV_, Feb 03 2014

%o (MAGMA) [Fibonacci(5*n)/5: n in [0..30]]; // _G. C. Greubel_, Dec 02 2017

%Y A column of array A028412.

%Y Cf. A000045, A102312, A243399.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_

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Last modified November 20 17:09 EST 2019. Contains 329337 sequences. (Running on oeis4.)