%I #11 Jun 13 2015 00:51:50
%S 1,2,24,160,1232,9120,68224,508928,3799296,28357120,211662848,
%T 1579868160,11792306176,88018952192,656982441984,4903783628800,
%U 36602339459072,273203580764160,2039219289063424,15220939987877888
%N Expansion of (1-4x)/(1-6x-12x^2+8x^3).
%C In general, sum{k=0..n, sum{j=0..n, C(2(n-k),j)C(2k,j)r^j}} has expansion (1-(r+1)x)/((1+(r+3)x+(r-1)(r+3)x^2+(r-1)^3*x^3).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,12,-8).
%F G.f.: (1-4x)/((1+2x)(1-8x+4x^2)); a(n)=6a(n-1)+12a(n-2)-8a(n-3); a(n)=sum{k=0..n, sum{j=0..n, C(2(n-k), j)C(2k, j)3^j}}.
%F Conjecture: a(n)=A002605(n+1)*A026150(n). [From _R. J. Mathar_, Jul 08 2009]
%F a(0)=1, a(1)=2, a(2)=24, a(n)=6*a(n-1)+12*a(n-2)-8*a(n-3) [From Harvey P. Dale, Feb 21 2012]
%F a(n) = (-2)^n/2 +A102591(n)/2. - _R. J. Mathar_, Sep 20 2012
%t CoefficientList[Series[(1-4x)/(1-6x-12x^2+8x^3),{x,0,30}],x] (* or *) LinearRecurrence[{6,12,-8},{1,2,24},30] (* _Harvey P. Dale_, Feb 21 2012 *)
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jun 04 2005