NOTES ON A214406 Peter Bala, Jul 25 2012 We extend results of Carlitz and Smiley connecting the tree function T(z) with the second-order Eulerian numbers to the case of the second-order Eulerian numbers of type B. --------------------------------------------------------------------------------------------------------------------------------- Suppose z = x*exp(-x). The inverse function x = T(z) is called the tree function. The Lagrange inversion formula gives the explicit expansion T(z) = Sum_{n >= 1} n^(n-1)*z^n/n!. Carlitz [2] considered exponential generating functions g(m,x) := Sum_{n>=1} n^(n+m)*z^n/n!, m an integer, so in particular g(-1,x) = T(z). He showed that for m >= 0, g(m,x) is a rational function in T(z) of the form T*P(m,T)/(1-T)^(2*m+1), where P(m,T) is a polynomial in T. The first few values are g(0,x) = T/(1 - T) g(1,x) = T/(1 - T)^3 g(2,x) = T*(1 + 2*T)/(1 - T)^5 g(3,x) = T*(1 + 8*T + 6*T^2)/(1 - T)^7. The numerator polynomials P(m,T) were later identified as the row generating polynomials of the second-order Eulerian numbers A008517. Smiley [3] extended Carlitz's result by proving that for m <= -1, g(m,x) is a polynomial in T(z). As we shall show, the approach used by Smiley can easily be extended to prove the corresponding results in the type B case. SECOND-ORDER EULERIAN NUMBERS OF TYPE B We introduce the type B analogues of the functions g(m,x). To this end, let now z = x*exp(-1/2*x^2). The Lagrange inversion formula gives x = Sum_{n >= 0} (2*n+1)^(n-1)*z^(2*n+1)/(2^n*n!). Since z^2 = x^2*exp(-x^2) we see that x may be expressed in terms of the tree function: x = sqrt(T(z^2)). Define functions F(m,z) and G(m,x) for integer values of m by F(m,z) := Sum_{n >= 0} (2*n+1)^(n+m-1)*z^(2*n+1)/(2^n*n!) G(m,x) := F(m,z) = Sum_{n >= 0} (2*n+1)^(n+m-1)*(x*exp(-x^2/2))^(2*n+1)/(2^n*n!). In particular, x = F(0,z) = G(0,x). Let R(n,x) denote the n-th row polynomial of A214406. In a previous note [1] we showed that the rational functions u(n,x) := x*R(n,x^2)/(1 - x^2)^(2*n+1) satisfied the recurrence u(n,x) = x/(1 - x^2)*d/dx(u(n-1,x)) (1) with starting value u(0,x) = x/(1 - x^2). The following proposition establishes the link between the functions G(m,x) and the row polynomials R(n,x). PROP0SITION G(n+1,x) = x*R(n,x^2)/(1 - x^2)^(2*n+1), n >= 0. PROOF For integer m G(m-1,x) = F(m-1,z) = Integral_{u = 0..z} F(m,u) du/u. Make the change of variable u = t*exp(-1/2*t^2), du/u = (1 - t^2)/t dt, to find G(m-1,x) = Integral_{t = 0..x} F(m,t*exp(-1/2*t^2)) * (1 - t^2)/t dt = Integral_{t = 0..x} G(m,t) * (1 - t^2)/t dt. Differentiating with respect to x gives d/dx(G(m-1,x)) = G(m,x)*(1 - x^2)/x and hence G(m,x) = x/(1 - x^2)*d/dx(G(m-1,x)). (2) Since G(0,x) = x we obtain G(1,x) = x/(1 - x^2). Comparing (1) with (2).we conclude that G(n+1,x) = u(n,x) = x*R(n,x^2)/(1 - x^2)^(2*n+1), n >= 0. END