%I #62 Sep 08 2022 08:44:58
%S 0,1,29,842,24447,709805,20608792,598364773,17373187209,504420793834,
%T 14645576208395,425226130837289,12346203370489776,358465123875040793,
%U 10407834795746672773,302185674200528551210,8773792386611074657863
%N a(n) = Fibonacci(7*n)/13.
%H G. C. Greubel, <a href="/A049667/b049667.txt">Table of n, a(n) for n = 0..650</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 (29, 1).
%F G.f.: x/(1 - 29*x - x^2).
%F a(n) = A134498(n)/13.
%F a(n) = F(n, 29), the n-th Fibonacci polynomial evaluated at x=29. - _T. D. Noe_, Jan 19 2006
%F a(n) = 29*a(n-1) + a(n-2), n > 1; a(0)=0, a(1)=1. - _Philippe Deléham_, Nov 22 2008
%F For n >= 1, a(n) equals the denominator of the continued fraction [29, 29, ..., 29] (with n copies of 29). The numerator of that continued fraction is a(n+1). - _Greg Dresden_ and _Shaoxiong Yuan_, Jul 26 2019
%F a(n) = ((-1)^n*7*F(n) + 14*5*F(n)^3 + (-1)^n*7*5^2*F(n)^5 + 5^3*F(n)^7)/13, n >= 0. See the general D. Jennings formula given in comment on triangle A111125, where also the reference is given. Here the fourth row (k=3) applies. - _Wolfdieter Lang_, Sep 01 2012
%F G.f.: G(0)*x/(2-29*x), where G(k)= 1 + 1/(1 - (x*(845*k-841))/((x*(845*k+4)) - 58/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 15 2013
%F O.g.f.: x*exp(Sum_{n >= 1} Lucas(7*n)*x^n/n) = x + 29*x^2 + 842*x^3 + .... - _Peter Bala_, Oct 11 2019
%p a:= n-> (<<0|1>, <1|29>>^n)[1, 2]:
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Sep 20 2017
%t Fibonacci[(7*Range[0,20])]/13 (* or *) LinearRecurrence[{29,1},{0,1},20] (* _Harvey P. Dale_, Sep 17 2017 *)
%o (MuPAD) numlib::fibonacci(7*n)/13 $ n = 0..25; // _Zerinvary Lajos_, May 09 2008
%o (Sage) [fibonacci(7*n)/13 for n in range(0, 17)] # _Zerinvary Lajos_, May 15 2009
%o (PARI) a(n)=fibonacci(7*n)/13 \\ _Charles R Greathouse IV_, Oct 07 2016
%o (Magma) [Fibonacci(7*n)/13: n in [0..30]]; // _G. C. Greubel_, Dec 02 2017
%Y A column of array A028412.
%Y Cf. A000045, A134498.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_