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%I #18 Dec 26 2022 08:40:27
%S 0,4,18,50,166,474,1478,4330,13206,39194,118438,353610,1064246,
%T 3185914,9571398,28686890,86115286,258236634,774928358,2324348170,
%U 6973918326,20920007354,62763517318,188283561450,564864665366,1694566034074
%N a(n) = (-4 + (-2)^n + 2*3^(n+1))/3 - [n=0].
%H G. C. Greubel, <a href="/A120656/b120656.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,5,-6).
%F From _Colin Barker_, Oct 19 2012: (Start)
%F a(n) = (-4 + (-2)^n + 2*3^(n+1))/3 - [n=0].
%F a(n) = 2*a(n-1) + 5*a(n-2) - 6*a(n-3) for n>3.
%F G.f.: 2*x*(2-x)*(1+3*x)/((1-x)*(1+2*x)*(1-3*x)). (End)
%F E.g.f.: (1/3)*(exp(-2*x) - 3 - 4*exp(x) + 6*exp(3*x)). - _G. C. Greubel_, Dec 25 2022
%t LinearRecurrence[{2,5,-6}, {0,4,18,50}, 51] (* _Harvey P. Dale_, Oct 18 2014 *)
%o (Magma) [0] cat [(-4 + (-2)^n + 2*3^(n+1))/3: n in [1..50]]; // _G. C. Greubel_, Dec 25 2022
%o (SageMath) [(-4 + (-2)^n + 2*3^(n+1))/3 -int(n==0) for n in range(51)] # _G. C. Greubel_, Dec 25 2022
%Y Cf. A120655.
%K nonn,less
%O 0,2
%A _Roger L. Bagula_, Aug 10 2006
%E Edited by _G. C. Greubel_, Dec 25 2022
%E Meaningful name using given formula from _Joerg Arndt_, Dec 26 2022
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