|
%I #28 May 11 2019 18:34:06
%S 0,2,16,512,34816,4063232,724566016,183240753152,62382319599616,
%T 27507470234550272,15250953398036463616,10384044045105304174592,
%U 8517992937742473694806016,8285310016381680852100186112
%N E.g.f. arctanh(tan(x)*tan(x)) (even powers only).
%C a(n) = 2*A000828(2*n-1). - corrected by _Vaclav Kotesovec_, Feb 08 2015
%F E.g.f.: log(sec(2*x))/2, cf. A000182. - _Vladeta Jovovic_, Jun 04 2005
%F a(n) = (-1)^(n(2n+1))4^(2n+1)(E{2n+1}(1/2)+E{2n+1}(1)); E_{n}(x) Euler polynomial. - _Peter Luschny_, Nov 25 2010
%F G.f.: 2*x/G(0) where G(k) = 1 - (2*k+2)*(2*k+4)*x/G(k+1)) (continued fraction, 1-step). - _Sergei N. Gladkovskii_, Oct 25 2012
%F G.f.: (2/G(0) - 1)*sqrt(-x), where G(k) = 1 + 1/(1 - 1/(1 + 1/(4*sqrt(-x)*(k+1)) - 1/G(k+1))) (continued fraction). - _Sergei N. Gladkovskii_, May 29 2013
%F G.f.: 2*x*T(0), where T(k) = 1 - x*(2*k+2)*(2*k+4)/(x*(2*k+2)*(2*k+4) - 1/T(k+1) ) (continued fraction). - _Sergei N. Gladkovskii_, Oct 12 2013
%F a(n) ~ 2^(4*n) * (2*n-1)! / Pi^(2*n). - _Vaclav Kotesovec_, Feb 08 2015
%e arctanh(tan(x)*tan(x)) = (2/2!)*x^2 + (16/4!)*x^4 + (512/6!)*x^6 + (34816/8!)*x^8 + ...
%p A012393 := n -> (-1)^(n*(2*n+1))*4^(2*n+1)*(euler(2*n+1,1/2) + euler(2*n+1,1)) end; # _Peter Luschny_, Nov 25 2010
%t nn = 20; Table[(CoefficientList[Series[ArcTanh[Tan[x]^2], {x, 0, 2*nn}], x] * Range[0, 2*nn]!)[[n]], {n, 1, 2*nn+1, 2}] (* _Vaclav Kotesovec_, Feb 08 2015 *)
%K nonn
%O 0,2
%A Patrick Demichel (patrick.demichel(AT)hp.com)
%E a(0)=0 prepended by _Joerg Arndt_, Oct 26 2012
| 