%I #28 Oct 14 2014 18:59:05
%S 0,1,-2,6,-20,80,-366,1904,-11080,71424,-505210,3891712,-32433180,
%T 290787328,-2791053734,28556359680,-310264194320,3567710830592,
%U -43287834157938,552688817143808,-7407423764750500,103981459115606016,-1525675236649033822,23354749389657604096
%N E.g.f. x/(1+sin(x)).
%F E.g.f. x/(1+sin(x)).
%F From Sergei N. Gladkovskii, May 29 2012: (Start)
%F Let E(x) be the e.g.f., then
%F E(x) = 1 - 1/(1+x)+ x^4/((1+x)*((1+x)*G(0) + x^2)) where G(k)= (2*k+2)*(2*k+3)-x^2+(2*k+2)*(2*k+3)*x^2/G(k+1); (continued fraction Euler's kind, 1-step ).
%F E(x) = 1 - 1/(1+x)+ x^4/((1+x)*((1+x)*G(0) + x^2)) where G(k)= 8*k+6-x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction Euler's 2 kind, 2-step ).
%F E(x) = x/(1+x*G(0)) where G(k)= 1 - x^2/(2*(2*k+1)*(4*k+3) - 2*x^2*(2*k+1)*(4*k+3)/(x^2 - 4*(k+1)*(4*k+5)/G(k+1))); (continued fraction 3 kind, 3-step ).
%F (End)
%F E.g.f. x/(1 + x*G(0)) where G(k)= 1 + x^2/( (2*k+1)*(2*k+3) - 2*k+1)*(2*k+3)^2/(2*k+3 - (2*k+2)/G(k+1))) ; (continued fraction, 3rd kind, 3-step). - _Sergei N. Gladkovskii_, Oct 01 2012
%t g[x_] = FullSimplify[x/(-1 + Sum[(-1)^n*x^(2*n - 1)/(2*n - 1)!, {n, 1, Infinity}])] h[x_, n_] = Dt[g[x], {x, n}] b[x_] = Table[h[x, n], {n, 0, 50}]; b[0]
%t With[{nn=30},CoefficientList[Series[x/(1+Sin[x]),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Oct 14 2014 *)
%o (Sage)
%o @CachedFunction
%o def c(n,k) :
%o if n==k: return 1
%o if k<1 or k>n: return 0
%o return ((n-k)//2+1)*c(n-1,k-1)+k*c(n-1,k+1)
%o def A108124(n): return (-1)^(n+1)*n*add(c(n,k) for k in (0..n))
%o [A108124(n) for n in (0..23)] # _Peter Luschny_, Jun 10 2014
%Y Cf. A186365
%K sign
%O 0,3
%A _Roger L. Bagula_, Jun 26 2005