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a(n) = (18^n - (-6)^n)/24.
0

%I #13 Dec 31 2023 16:24:45

%S 0,1,12,252,4320,79056,1415232,25520832,459095040,8265390336,

%T 148766948352,2677865536512,48201216860160,867624080265216,

%U 15617220384079872,281110045277601792,5059980344811847680,91079649027723165696,1639433665572357537792,29509806081862392348672

%N a(n) = (18^n - (-6)^n)/24.

%C In general k^(n-1)*A015518(n) is given by ((3k)^n-(-k)^n)/(4k) with g.f. x/((1+kx)(1-3kx)).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,108).

%F G.f.: x/((1+6x)*(1-18x)).

%F a(n) = 12a(n-1)+108a(n-2). a(n) = 6^(n-1)*A015518(n).

%t LinearRecurrence[{12,108},{0,1},20] (* _Harvey P. Dale_, May 24 2015 *)

%Y Cf. A053524, A053535, A053536, A053537.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Sep 29 2004