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A180168 a(n) = 2*a(n-1) + 5*a(n-2), a(0) = 1, a(1) = 3.

%I A180168 #7 Feb 18 2017 22:07:03
%S A180168 1,3,11,37,129,443,1531,5277,18209,62803,216651,747317,2577889,
%T A180168 8892363,30674171,105810157,364991169,1259033123,4343022091,
%U A180168 14981209797,51677530049,178261109083,614909868411,2121125282237,7316799906529,25239226224243,87062451981131
%N A180168 a(n) = 2*a(n-1) + 5*a(n-2), a(0) = 1, a(1) = 3.
%H A180168 G. C. Greubel, <a href="/A180168/b180168.txt">Table of n, a(n) for n = 0..1000</a>
%H A180168 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,5).
%F A180168 G.f.: (1 + x)/(1 - 2*x - 5*x^2).
%F A180168 Equals INVERT transform of A026597: (1, 2, 6, 14, 38, 94,...).
%F A180168 a(n) = (1/6)*( -(1-sqrt(6))^n*sqrt(6) + sqrt(6)*(1+sqrt(6))^n + 3*(1-sqrt(6))^n + 3*(1 +sqrt(6))^n ). - _Alexander R. Povolotsky_, Aug 15 2010
%F A180168 a(n) = A176812(n)/3 = A002532(n) + A002532(n+1). - R. J. Mathar, Oct 11 2011
%e A180168 a(5) = 443 = 2*a(4) + 5*a(3) = 2*129 + 5*37.
%e A180168 Using the INVERT operation, a(4) = 129 = (38, 14, 6, 2, 1) dot (1, 1, 3, 11, 37)
%e A180168 = (38 + 14 + 18 + 22 + 37); where A026597 = (1, 2, 6, 14, 38, 94,...).
%t A180168 LinearRecurrence[{2, 5}, {1, 3}, 50] (* _G. C. Greubel_, Feb 18 2017 *)
%o A180168 (PARI) x='x+O('x^25); Vec((1 + x)/(1 - 2*x - 5*x^2)) \\ _G. C. Greubel_, Feb 18 2017
%Y A180168 Cf. A026597.
%K A180168 nonn,easy
%O A180168 0,2
%A A180168 _Gary W. Adamson_, Aug 14 2010