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.

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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). 

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 analogs 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)
	
satisified 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) = int {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) = int {u = 0..x} F(m,t*exp(-1/2*t^2)) * (1-t^2)/t dt

		 = int {u = 0..x} G(m,t)*(1-t^2)/t dt.

Diffentiating 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) = x*R(n,x^2)/(1-x^2)^(2*n+1),   n >= 0.

END


The rational functions R(n,x)/(1-x)^(2*n+1) are the o.g.f.'s for the diagonals of A039755 - the type B Stirling numbers of the second

kind. Proposition 1 thus gives alternative generating functions for these diagonal sequences in terms of the functions G(n,x).

We can also use (2) to compute G(m,x) for m a negative integer. A simple induction shows that the result will be a polynomial in x

(of degree 2*m+1, vanishing at x = 0). The first few values of G(m,x) (= F(m,z)) are tabulated below.


	Table of F(m,z), -4 <= m <= 4; x = F(0,z)
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	F(4,z) = x*(1 + 33*x^2 + 71*x^4 + 15*x^6)/(1 - x^2)^7

	F(3,z) = x*(1 + 8*x^2 + 3*x^4)/(1 - x^2)^5

	F(2,z) = x*(1 + x^2)/(1 - x^2)^3 

	F(1,z) = x/(1 - x^2)

	F(0,z) = x

	F(-1,z) = x*(1 - x^2/3 )

	F(-2,z) = x*(1 - 4*x^2/9 + x^4/15)

	F(-3,z) = x*(1 - 13*x^2/27 + 23*x^4/225 - x^6/105)

	F(-4,z) = x*(1 - 40*x^2/81 + 394*x^4/3375 - 176*x^6/11025 + x^8/945)



REFERENCES

[1] P. Bala, 		<a href="/A214406/a214406.txt">Second-order Eulerian numbers of type B</a>

[2] L. Carlitz,   	The Coefficients in an Asymptotic Expansion, Proc. Amer. Math. Soc., 16, (1965), 248-252 

[3] L. M. Smiley, 	Completion of a rational function sequence of Carlitz. arXiv:math/0006106v1 [math.CO]

	      		- available online at http://arxiv.org/abs/math.CO/0006106

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