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A081335 a(n) = (6^n + 2^n)/2.

%I A081335 #29 Sep 08 2022 08:45:09
%S A081335 1,4,20,112,656,3904,23360,140032,839936,5039104,30233600,181399552,
%T A081335 1088393216,6530351104,39182090240,235092508672,1410554986496,
%U A081335 8463329787904,50779978465280,304679870267392,1828079220555776
%N A081335 a(n) = (6^n + 2^n)/2.
%C A081335 Binomial transform of A034478. 4th binomial transform of (1, 0, 4, 0, 16, 0, 64, ...).
%C A081335 Case k=4 of the family of recurrences a(n) = 2*k*a(n-1) - (k^2-4)*a(n-2), a(0)=1, a(1)=k.
%H A081335 Vincenzo Librandi, <a href="/A081335/b081335.txt">Table of n, a(n) for n = 0..200</a>
%H A081335 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-12).
%F A081335 a(n) = 8*a(n-1) - 12*a(n-2), a(0)=1, a(1)=4.
%F A081335 G.f.: (1-4*x)/((1-2*x)*(1-6*x)).
%F A081335 E.g.f.: exp(4*x)*cosh(2*x).
%F A081335 a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * 4^(n-k) = Sum_{k=0..n} binomial(n,k) * 4^(n-k/2) * (1+(-1)^k)/2. - _Paul Barry_, Nov 22 2003
%F A081335 a(n) = Sum_{k=0..n} 4^k*A098158(n,k). - _Philippe Deléham_, Dec 04 2006
%t A081335 LinearRecurrence[{8, -12}, {1, 4}, 30] (* _Harvey P. Dale_, May 03 2013 *)
%t A081335 CoefficientList[Series[(1-4x)/((1-2x)(1-6x)), {x,0,30}], x] (* _Vincenzo Librandi_, Aug 08 2013 *)
%o A081335 (Magma) [(6^n+2^n)/2: n in [0..30]]; // _Vincenzo Librandi_, Aug 08 2013
%o A081335 (PARI) a(n)=(6^n+2^n)/2 \\ _Charles R Greathouse IV_, Oct 07 2015
%o A081335 (Sage) [2^(n-1)*(3^n + 1) for n in (0..30)] # _G. C. Greubel_, Aug 02 2019
%o A081335 (GAP) List([0..30], n-> 2^(n-1)*(3^n + 1)); # _G. C. Greubel_, Aug 02 2019
%Y A081335 Cf. A081336.
%K A081335 nonn,easy
%O A081335 0,2
%A A081335 _Paul Barry_, Mar 18 2003