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%I #16 Jul 02 2017 10:45:39
%S 0,1,2,6,28,160,1086,8624,78296,799488,9070810,113208832,1541351604,
%T 22734473216,361121134934,6145880954880,111569141960752,
%U 2151953994809344,43948641637067058,947412315736506368
%N E.g.f.: x/(1-sinh(x)).
%C "Bernoulli numbers" for x/(1-sinh(x)).
%F E.g.f. x/(1-sinh(x)).
%F From _Sergei N. Gladkovskii_, May 30 2012: (Start)
%F Let E(x)=x/(1-sinh(x)) be the e.g.f., then
%F E(x)=- 1 + 1/(1-x)+ x^4/((1-x)*((1-x)*G(0) - x^2)) ; 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)) ; G(k)= 8*k+6+x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction, Euler's 2nd kind, 2-step).
%F E(x)= x/(1 - x*G(0)); 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).
%F (End)
%F a(n) ~ n!/(sqrt(2)*(log(1+sqrt(2)))^n). - _Vaclav Kotesovec_, Jun 27 2013
%p G:=x/(1-sinh(x)): Gser:=series(G,x=0,25): 0,seq(n!*coeff(Gser,x^n),n=1..22);
%t g[x_] = x/(-1 + Sinh[x]) h[x_, n_] = Dt[g[x], {x, n}] a[x_] = Table[ -h[x, n], {n, 0, 50}]; b = a[0]
%t With[{nn=20},CoefficientList[Series[x/(1-Sinh[x]),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Jul 02 2017 *)
%K nonn
%O 0,3
%A _Roger L. Bagula_, Jun 27 2005
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