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%I #17 Mar 09 2014 04:49:07
%S 1,0,-4,24,-96,0,5760,-80640,645120,0,-116121600,2554675200,
%T -30656102400,0,11158821273600,-334764638208000,5356234211328000,0,
%U -3278015337332736000,124564582818643968000,-2491291656372879360000,0
%N Expansion of tanh(log(1+1/x)).
%F E.g.f.: (2*x+1)/(2*x^2+2*x+1) = 1-4*x^2/2!+24*x^3/3!-96*x^4/4!+....
%F Recurrence: a(n) = -2*n*a(n-1)-2*n*(n-1)*a(n-2), a(0) = 1, a(1) = 0.
%F a(n) = -n!/2*((-1+i)^(n+1) + (-1-i)^(n+1)) = -n!*sqrt(2)^(n+1)* cos(3*Pi*(n+1)/4).
%F a(n) = 2^n*A009014(n). a(n) = -n!*A009116(n+1).
%F For x > -1/2 we have (2*x+1)/(2*x^2+2*x+1) = 2*int {t = 0..inf} exp(-t*(2*x+1))*cos(t). Using this we obtain a(n) = 2*(-2)^n*int {t = 0..inf} t^n*exp(-t)*cos(t). - _Peter Bala_, Oct 05 2011
%t nn = 26; Range[0, nn]! CoefficientList[Series[Tanh[Log[1 + 1/x]], {x, 0, nn}], x] (* _T. D. Noe_, Oct 05 2011 *)
%Y Cf. A009014, A009116.
%K sign,easy
%O 0,3
%A _R. H. Hardin_
%E Extended with signs by _Olivier Gérard_, Mar 15 1997