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A099012 a(n) = 3^(n-1)*Fibonacci(n).

%I A099012 #38 Sep 08 2022 08:45:15
%S A099012 0,1,3,18,81,405,1944,9477,45927,223074,1082565,5255361,25509168,
%T A099012 123825753,601059771,2917611090,14162371209,68745613437,333698181192,
%U A099012 1619805064509,7862698824255,38166342053346,185263315578333
%N A099012 a(n) = 3^(n-1)*Fibonacci(n).
%C A099012 Binomial transform is A057088 (with leading 0). Partial sums are A099013. Binomial transform of A015447 (with leading 0).
%C A099012 The ratio a(n+1)/a(n) converges to 3 times the golden ratio (of A000045) as n approaches infinity. - _Felix P. Muga II_, Mar 10 2014
%H A099012 G. C. Greubel, <a href="/A099012/b099012.txt">Table of n, a(n) for n = 0..1000</a>
%H A099012 F. P. Muga II, <a href="https://www.researchgate.net/publication/267327689_Extending_the_Golden_Ratio_and_the_Binet-de_Moivre_Formula">Extending the Golden Ratio and the Binet-de Moivre Formula</a>, March 2014.
%H A099012 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,9).
%F A099012 G.f.: x/(1 - 3*x - 9*x^2).
%F A099012 a(n) = 3*a(n-1) + 9*a(n-2).
%F A099012 a(n) = sqrt(5)(3/2 + 3*sqrt(5)/2)^n/15 - sqrt(5)*(3/2 - 3*sqrt(5)/2)^n/15.
%t A099012 a[n_]:=(MatrixPower[{{1,5},{1,-4}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 20 2010 *)
%t A099012 Table[3^(n-1) Fibonacci[n],{n,0,30}] (* or *) LinearRecurrence[{3,9},{0,1},30] (* _Harvey P. Dale_, Nov 07 2017 *)
%o A099012 (Sage) [lucas_number1(n,3,-9) for n in range(0, 23)] # _Zerinvary Lajos_, Apr 22 2009
%o A099012 (Magma) [3^(n-1)*Fibonacci(n): n in [0..60]]; // _Vincenzo Librandi_, Apr 23 2011
%o A099012 (PARI) a(n)=3^(n-1)*fibonacci(n) \\ _Charles R Greathouse IV_, Sep 24 2015
%Y A099012 Cf. A000045, A057088, A099013, A015447.
%K A099012 easy,nonn
%O A099012 0,3
%A A099012 _Paul Barry_, Sep 22 2004