Let b(n)=(2n)!*(2n+2)!/(2*n!*(n+1)!^2*(n+2)!). Let B(x) be the generating function producing b(n), and A(x) be the generating function producing a(n). Then these sequences satisfy the functional equation B(x)=A(x(1+2*B(x))^2). - Sean A. Irvine, Apr 05 2010
MATHEMATICA
max = 21; b[n_] := (2n)!*(2n + 2)!/(2*n!*(n + 1)!^2*(n + 2)!); b[0] = 0; bf[x_] := Sum[b[n]*x^n, {n, 0, max}]; Clear[a]; a[0] = 0; a[1] = a[2] = 1; af[x_] := Sum[a[n]*x^n, {n, 0, max}]; se = Series[bf[x] - af[x*(1 + 2*bf[x])^2], {x, 0, max}] // Normal; Table[a[n], {n, 1, max}] /. SolveAlways[se == 0, x] // First (* Jean-François Alcover, Jan 31 2013, after Sean A. Irvine *)