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a(n) = 0^n/2 + 2^n*(n^2+n+2)/4.
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%I #15 Jan 02 2021 12:35:46

%S 1,2,8,28,88,256,704,1856,4736,11776,28672,68608,161792,376832,868352,

%T 1982464,4489216,10092544,22544384,50069504,110624768,243269632,

%U 532676608,1161822208,2524971008,5469372416,11811160064,25434259456

%N a(n) = 0^n/2 + 2^n*(n^2+n+2)/4.

%C Binomial transform of A080335 (with additional leading 1).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,8).

%F a(n) = 2*A007466(n) for n >= 1.

%F G.f.: (1-4*x+8*x^2-4*x^3)/(1-2*x)^3. - _Colin Barker_, Mar 18 2012

%t LinearRecurrence[{6,-12,8},{1,2,8,28},30] (* _Harvey P. Dale_, Nov 26 2015 *)

%K easy,nonn

%O 0,2

%A _Paul Barry_, Sep 02 2003