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a(n) = 3^n - Fibonacci(n).
3

%I #16 Dec 08 2023 13:40:58

%S 1,2,8,25,78,238,721,2174,6540,19649,58994,177058,531297,1594090,

%T 4782592,14348297,43045734,129138566,387417905,1162257286,3486777636,

%U 10460342257,31381041898,94143150170,282429490113,847288534418,2541865706936,7625597288569

%N a(n) = 3^n - Fibonacci(n).

%H Bruno Berselli, <a href="/A212323/b212323.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.: (1 - 2*x + 2*x^2)/((1 - 3*x)*(1 - x - x^2)).

%t Table[3^n - Fibonacci[n], {n, 0, 27}]

%t LinearRecurrence[{4,-2,-3},{1,2,8},30] (* _Harvey P. Dale_, Dec 08 2023 *)

%o (PARI) for(n=0, 27, print1(3^n-fibonacci(n)", "));

%o (Magma) [3^n-Fibonacci(n): n in [0..27]];

%Y Cf. A000045, A000244, A099036, A212262.

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, May 09 2012