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%I #36 Oct 28 2022 17:10:36
%S 0,2,16,272,7936,353792,22368256,1903757312,209865342976,
%T 29088885112832,4951498053124096,1015423886506852352,
%U 246921480190207983616,70251601603943959887872,23119184187809597841473536,8713962757125169296170811392,3729407703720529571097509625856
%N Tan(x)^2 = sum(n>=0, a(n)*x^(2*n)/(2*n)! ).
%F (tan(z))^2 = z^2/(1-z^2)*( 1 +2*z^2/( (z^2-1)*(G(0)-2*z^2)), G(k) = (k+2)*(2*k+3)-2*z^2+2*z^2*(k+2)*(2*k+3)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Dec 15 2011
%F (tan(z))^2 = z^2/(G(0)+z^2) where G(k) = (k+1)*(2*k+1)-2*z^2+2*z^2*(k+1)*(2*k+1)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Dec 15 2011
%F G.f. A(x)=-1 + 1/G(0) where G(k)= 1 - (k+1)*(k+2)*x/G(k+1); (continued fraction, 1-step). - _Sergei N. Gladkovskii_, Aug 10 2012
%F G.f.: 1/G(0)-1 where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Jan 13 2013
%F G.f.: (1/G(0)-1)*sqrt(-x), where G(k)= 1 - sqrt(-x) - x*(k+1)^2/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, May 29 2013
%F G.f.: Q(0) -1, where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Oct 14 2013
%e (tan x)^2 = x^2 + 2/3*x^4 + 17/45*x^6 + 62/315*x^8 + ...
%t With[{nn=30},Take[CoefficientList[Series[Tan[x]^2,{x,0,nn}],x] Range[0,nn]!, {1,-1,2}]] (* _Harvey P. Dale_, Oct 04 2011 *)
%Y Essentially same as A000182.
%Y Cf. A024283, A000182.
%K nonn,easy
%O 0,2
%A _R. H. Hardin_
%E Extended and signs tested Mar 15 1997 by _Olivier Gérard_.
%E More terms from _Harvey P. Dale_, Oct 04 2011
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