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%I #31 Feb 09 2023 19:28:42
%S 0,1,4,14,45,140,428,1297,3912,11770,35365,106184,318696,956321,
%T 2869340,8608630,25826877,77482228,232449268,697351985,2092062720,
%U 6276199106,18828615029,56485873744,169457667600,508373077825,1525119354868
%N Convolution of Fibonacci(n) and 3^n.
%H Vincenzo Librandi, <a href="/A094688/b094688.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-3).
%F G.f.: x/((1-3*x)*(1-x-x^2)).
%F a(n) = (1/5)*(3^(n+1) - Lucas(n+2)).
%F a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
%F a(n) = A101220(3, 3, n). - _Ross La Haye_, Jan 28 2005
%F a(n) = a(n-1) + a(n-2) + 3^(n-1) for n > 1, with a(0) = 0, a(1) = 1. - _Ross La Haye_, Aug 20 2005
%F a(n) = 3*a(n-1) + Fibonacci(n), where a(0) = 0. - _Taras Goy_, Mar 24 2019
%t LinearRecurrence[{4,-2,-3},{0,1,4},40] (* _Vincenzo Librandi_, Jun 24 2012 *)
%t Table[(3^(n+1) -LucasL[n+2])/5, {n,0,40}] (* _Vladimir Reshetnikov_, Sep 27 2016 *)
%o (PARI) a(n)=(3^(n+1)-fibonacci(n+1)-fibonacci(n+3))/5 \\ _Charles R Greathouse IV_, Jun 28 2011
%o (Magma) I:=[0,1,4]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2) -3*Self(n-3): n in [1..41]]; // _Vincenzo Librandi_, Jun 24 2012
%o (SageMath) [(3^(n+1) -lucas_number2(n+2,1,-1))/5 for n in range(41)] # _G. C. Greubel_, Feb 09 2023
%Y Cf. A000032, A000045, A101220.
%K easy,nonn
%O 0,3
%A _Paul Barry_, May 19 2004
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