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a(n) = (n+2)*2^(n-1) - 2*n.
2

%I #18 Jul 23 2019 08:11:18

%S 1,1,4,14,40,102,244,562,1264,2798,6124,13290,28648,61414,131044,

%T 278498,589792,1245150,2621404,5504986,11534296,24117206,50331604,

%U 104857554,218103760,452984782,939524044,1946157002,4026531784

%N a(n) = (n+2)*2^(n-1) - 2*n.

%H Harry J. Smith, <a href="/A066368/b066368.txt">Table of n, a(n) for n = 0..200</a>

%H M. Azaola and F. Santos, <a href="http://personales.unican.es/santosf/Articulos/">The number of triangulations of the cyclic polytope C(n,n-4)</a>, Discrete Comput. Geom., 27 (2002), 29-48.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6,-13,12,-4).

%F G.f.: (1-5*x+11*x^2-9*x^3)/((1-x)^2*(1-2*x)^2). - _Colin Barker_, Apr 19 2012

%t Table[(n+2)2^(n-1)-2n,{n,0,30}] (* _Harvey P. Dale_, Apr 18 2019 *)

%o (PARI) { for (n=0, 200, write("b066368.txt", n, " ", (n + 2)*2^(n - 1) - 2*n) ) } \\ _Harry J. Smith_, Feb 11 2010

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jan 04 2002