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a(n) = (1/6)*(3^n + (-2)^n - 1).
0

%I #37 Jun 20 2021 03:05:43

%S 0,2,3,16,35,132,343,1136,3195,10012,29183,89256,264355,799892,

%T 2386023,7185376,21501515,64613772,193622863,581305496,1743042675,

%U 5230875652,15689131703,47074385616,141209175835,423655489532,1270910544543,3812843481736,11438306748995

%N a(n) = (1/6)*(3^n + (-2)^n - 1).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,5,-6).

%F G.f.: x*(2 - x)/((1 - x)*(1 + 2*x)*(1 - 3*x)). - _Andrew Howroyd_, Jun 15 2021

%F a(n) = A094554(n-1) + 2*A094556(n-1). - _Greg Dresden_, Jun 19 2021

%e a(4) = (1/6)*(3^4 + (-2)^4 - 1) = (1/6)*(81+16-1) = 16.

%t LinearRecurrence[{2, 5, -6}, {0, 2, 3}, 30] (* _Greg Dresden_, Jun 19 2021 *)

%Y Potentially related to A094554, A094555, and A094556 (which have the same recurrence).

%K nonn,easy

%O 1,2

%A _Ryan Brooks_, Jun 14 2021