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%I #3 Mar 30 2012 18:36:51
%S 0,0,0,1,-2,4,-7,13,-24,48,-99,221,-512,1268,-3247,8773,-24400,70896,
%T -211347,653541,-2068472,6755684,-22541135,77305981,-270435640,
%U 969413776,-3539893923,13212871629,-50180362320,194412817844,-765590169935,3070433223317
%N a(n) = A113290(n,1)/(n+1) for n>=0, where A113290 is the matrix log of triangle A113287.
%F G.f. satisfies: A(x) = x^3*((2+x)/(1+x) + (1+x)*A'(x))/(2+3*x+2*x^2). a(n+3) = (-1)^n*Sum_{k=0..n} Sum{j=0..[k/2]} (k-j)!/(k-2*j)! for n>=0. a(n+3) = -a(n+2) + (-1)^n*A072374(n) for n>=1.
%o (PARI) a(n)=if(n<3,0,(-1)^(n-3)*sum(k=0,n-3,sum(j=0,k\2,(k-j)!/(k-2*j)!)))
%Y Cf. A113287, A113288, A113290, A072374.
%K sign
%O 0,5
%A _Paul D. Hanna_, Oct 23 2005
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