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Expansion of f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3).
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%I #21 Nov 12 2018 18:56:23

%S 1,0,-1,-1,2,5,-2,-21,-14,72,138,-165,-803,-143,3575,4732,-11674,

%T -36244,15130,195738,152456,-799102,-1700272,2042975,11038183,2582670,

%U -53547795,-76684530,185864265,618689190,-231325605,-3506922585,-2974386450,14866619160,33459332610,-38401746930,-223156727472

%N Expansion of f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3).

%C The compositional inverse F^{[-1]} of F(x) = x/(1 - x^2 - x^3) (appearing in the Riordan matrix of the Bell type R = (F(x)/x, F(x)) from A104578) is needed for the inverse matrix of R, the Riordan matrix R^(-1) = (f(t), t*f(t)).

%C This function f(t) is also needed for the A- and Z-sequences of this Riordan matrix R, namely A(n) = [t^n](1/f(t)) = A319202(n) and Z(n) = [t^n]((1/t)*(1/f(t) - 1)) = A319202(n+1).

%C For the expansion of the compositional inverse of x/(1 + x^2 + x^3) see A001005.

%F a(n) = [t^n] G(t), where G(t) = F^{[-1]}(t)/t, and F(x) = x/(1 - x^2 - x^3).

%F O.g.f.: G(t) can be computed by Lagrange inversion of F.

%F a(n) = b(n+1)/(n+1), for n >= 0, where b(n+1) is the coefficient of x^n of (1 - x^2 - x^3)^(n+1). This follows from the Lagrange inversion series for G(x) = F^{[-1]}(x)/x.

%F a(n) = (1/(n+1))*(Sum_{2*e2 + 3*e3 = n} (-1)^{e2 + e3}*(n+1)!/(n+1 - (e2 + e3))!*e2!*e3!) (from the multinomial formula for (x1 + x2 + x3)^(n+1)) with x1 = 1, x2 = -x^2 and x3 = -x^3. For the solutions of 2*e2 + 3*e3 = n >= 2, and the parity of e2 + e3, see the array A321201.

%e a(8) = (1/9)*[x^8] (1- x^2 - x^3)^9 = (1/9)*(-126) = -14.

%e a(8) = (1/9)*(- 9!/(6!*1!*2!) + 9!/(5!*0!*4!)) = -14, from the two solutions for [e2, e3], namely [1, 2] (parity odd) and [0, 4] (parity even).

%p f := series(x/(1 - x^2 -x^3), x, 40):

%p r := gfun:-seriestoseries(f, 'revogf'):

%p gf := convert(r, polynom) / x:

%p seq(coeff(gf,x,n), n=0..36); # _Peter Luschny_, Nov 09 2018

%Y Cf. A001005, A104578, A319202.

%K sign

%O 0,5

%A _Wolfdieter Lang_, Nov 06 2018