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A090040 (3*6^n + 2^n)/4.

%I A090040 #32 Sep 08 2022 08:45:12
%S A090040 1,5,28,164,976,5840,35008,209984,1259776,7558400,45349888,272098304,
%T A090040 1632587776,9795522560,58773127168,352638746624,2115832446976,
%U A090040 12694994616320,76169967566848,457019805138944,2742118830309376
%N A090040 (3*6^n + 2^n)/4.
%C A090040 A090040 is the Q-residue of the triangle A175840, where Q is the triangular array (t(i,j)) given by t(i,j)=1; see A193649 for the definition of Q-residue. - _Clark Kimberling_, Aug 07 2011
%H A090040 Vincenzo Librandi, <a href="/A090040/b090040.txt">Table of n, a(n) for n = 0..300</a>
%H A090040 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-12).
%F A090040 G.f.: (1-3*x)/((1-2*x)*(1-6*x)).
%F A090040 E.g.f.: (3*exp(6*x)+exp(2*x))/4 = exp(4*x)*(cosh(2*x)+sinh(2*x)/2).
%F A090040 a(n) = 8*a(n-1) -12*a(n-2), a(0)=1, a(1)=5.
%F A090040 a(n) = (3*6^n+2^n)/4.
%F A090040 a(n)=6*a(n-1)-2^(n-1). - _Paul Curtz_, Jan 09 2009
%F A090040 Fourth binomial transform of (1, 1, 4, 4, 16, 16, ...). a(n)=sum{k=1..floor(n/2), C(n, 2k)4^(n-k-1)}. - _Paul Barry_, Nov 22 2003
%F A090040 a(n) = A019590 (mod 4), proof via a(n)=8*a(n-1)-12*a(n-2). - _R. J. Mathar_, Feb 25 2009
%F A090040 a(n) = Sum_{k, 0<=k<=n} A117317(n,k)*3^k. - _Philippe Deléham_, Jan 28 2012
%t A090040 LinearRecurrence[{8,-12},{1,5},30] (* _Harvey P. Dale_, Nov 23 2014 *)
%o A090040 (Magma) [(3*6^n+2^n)/4: n in [0..30]]; // _Vincenzo Librandi_, Jun 10 2011
%o A090040 (PARI) a(n)=(3*6^n+2^n)/4 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y A090040 Cf. A081335.
%K A090040 easy,nonn
%O A090040 0,2
%A A090040 _Paul Barry_, Nov 20 2003