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A206305
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E.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n+1)/(2*n+1)! is inverse to f(x) = 2*x - tan(x).
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1
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1, 2, 56, 4304, 647552, 161009408, 59798825984, 31018594543616, 21421107883900928, 19000925396453752832, 21053097631093130264576, 28496291901064818145624064, 46267687031143770070546644992, 88754774204916230754873248841728, 198588172996269363811346709494104064
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OFFSET
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0,2
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LINKS
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F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
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FORMULA
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a(n) = (sum(k=1..2*n, (2*n+k)!*sum(j=1..k, (-1)^(j)/(k-j)!*((sum(l=0..j-1, (1/((j-l)!*l!)*sum(m=j-l..2*n-l+j, binomial(m-1,-l+j-1)*m!*2^(2*n-m+j)*(-1)^(n+m)*Stirling2(2*n-l+j,m)))/(2*n-l+j)!)))))), n>0, a(0)=1.
From Peter Bala, Aug 06 2012: (Start)
Alternative form of the o.g.f: x/2 - (1/2)*(x - 2*atan(x))^<-1> = x/2 - (1/2)*(Integral_{t = 0..x} 1/f(t) dt)^<-1>, where f(x) = (x^2+1)/(x^2-1) and <-1> denotes series reversion w.r.t. x.
Applying Theorem 4.1 of Dominici to invert the integral gives the result: for n >= 1, a(n) = (1/2)*D^(2*n+1)(x) evaluated at x = 0, where D is the operator (1+x^2)/(1-x^2)*d/dx.
Applying Theorem 1 of Bergeron et al. gives an interpretation for the sequence as counting a variety of trees defined by the degree function phi(x) = (1+x^2)/(1-x^2) = 1 + 2*x^2 + 2*x^4 + .... We find for n >= 1 that 2*a(n) gives the number of rooted increasing plane (ordered) trees on (2*n+1) vertices with no vertices of odd outdegree and where the vertices of even outdegree come in two colors.
(End)
a(n) ~ 2^(2*n) * n^(2*n) / (exp(2*n) * (Pi/2-1)^(2*n+1/2)). - Vaclav Kotesovec, Jan 26 2014
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MATHEMATICA
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nmax = 11; coes = CoefficientList[ x/2 - (1/2)*InverseSeries[ Series[(x - 2*ArcTan[x]), {x, 0, 2*nmax+2}], x], x]*Range[0, 2*nmax+1]!; a[n_] := coes[[2*n+2]]; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Sep 05 2012 *)
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PROG
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(Maxima) a(n):=if n=0 then 1 else (sum((2*n+k)!*sum((-1)^(j)/(k-j)!*((sum((1/((j-l)!*l!)*sum(binomial(m-1, -l+j-1)*m!*2^(2*n-m+j)*(-1)^(n+m)*stirling2(2*n-l+j, m), m, j-l, 2*n-l+j))/(2*n-l+j)!, l, 0, j-1))), j, 1, k), k, 1, 2*n));
(PARI) seq(n)={my(v=Vec(serlaplace(serreverse(2*x - tan(x + O(x^(2*n+2))))))); vector(n+1, i, v[2*i-1])} \\ Andrew Howroyd, Apr 28 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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