a(0) = 0, a(4n+3) = 0, a(n) = (-1)^[n == 2, 5, 8 mod 8] * n!/2^floor(n/2). -
Ralf Stephan, Mar 06 2004
(1): a(n) = i*n!/2^(n+1)*{(i-1)^(n+1)-(-1-i)^(n+1)} for n>=1.
The function tanh(log(1+x)) is a disguised form of the rational function (x^2+2*x)/(x^2+2*x+2). Observe that
(2): (x^2+2*x)/(x^2+2*x+2) = d/dx[x - atan((x^2+2*x)/(2*x+2))].
Hence, with an offset of 1, the egf for this sequence is
(3): x - atan((x^2+2*x)/(2*x+2)) = x^2/2! - x^3/3! + 6*x^5/5!- 30*x^6/6! + 90*x^7/7! - ....
This sequence is closely related to the series reversion of the function E(x)-1, where E(x) = sec(x)+tan(x) is the egf for the sequence of zigzag numbers
A000111. Under the change of variable x -> sec(x)+tan(x)-1 the rational function (x^2+2*x)/(2*x+2) transforms to tan(x). Hence atan((x^2+2*x)/(2*x+2)) is the inverse function of sec(x)+tan(x)-1.
Recurrence relation:
(4): 2*a(n)+2*n*a(n-1)+n*(n-1)*a(n-2) = 0 with a(1) = 1, a(2) = -1.
(End)
