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FORMULA
| a(m, k)=0 if m<k+1; a(1, 0)=-1; a(m, -1):= 0; a(m, k) = -(2*m-k-1)*(a(m-1, k) + a(m-1, k-1)) else.
Contribution from Tom Copeland, May 05 2010 (Updated Sep 12 2011): (Start)
The integral from 0 to infinity w.r.t. w of
exp[-w(u+1)] (1+u*z*w)^(1/z) gives a power series, f(u,z), in z for reversed row polynomials in u of A111999, related to an Euler transform of diagonals of A008275.
Let g(u,x) be obtained from f(u,z) by replacing z^n with x^(n+1)/(n+1)!；
g(u,x)= x - u^2 x^2/2! + (2 u^2 + 3 u^4) x^3/3! - (6 u^4 + 20 u^5 + 15 u^6) x^4/4! + ... , an e.g.f. associated to f(u,z).
Then g^(-1)(u,x)=(1+u)*x - ln(1+u*x) is the comp. inverse of g(u,x) in x, and, consequently, A133932 is a refinement of A111999.
With h(u,x)= 1/(dg^(-1)/dx)= (1+u*x)/(1+(1+u)*u*x),
g(u,x)=exp[x*h(u,t)d/dt] t , evaluated at t=0. Also, dg(u,x)/dx = h(u,g(u,x)).
(End)
Contribution from Tom Copeland, May 06 2010: (Start)
For m,k>0, a(m,k) = Sum(j=2 to 2m-k+1): (-1)^(2m-k+1+j) C(2m-k+1,j) St1d(j,m) ,
where C(n,j) is the binomial coefficient and St1d(j,m) is the (j-m)-th element
of the m-th subdiagonal of A008275 for (j-m)>0 and is 0 otherwise,
e.g., St1d(1,1) = 0, St1d(2,1) = -1, St1d(3,1) = -3, St1d(4,1) = -6.
(End)
Contribution from Tom Copeland, Sep 03 2011 (Updated Sep 12 2011): (Start)
The integral from 0 to infinity w.r.t. w of
exp[-w*(u+1)/u] (1+u*z*w)^(1/(u^2*z)) gives a power series, F(u,z), in z for the row polynomials in u of A111999.
Let G(u,x) be obtained from F(u,z) by replacing z^n with x^(n+1)/(n+1)!;
G(u,x) = x - x^2/2! + (3 + 2 u) x^3/3! - (15 + 20 u + 6 u^2) x^4/4! + ... , an e.g.f. for A111999 associated to F(u,z).
G^(-1)(u,x)=((1+u)*u*x - ln(1+u*x))/u^2 is the comp. inverse of G(u,x)in x.
With H(u,x) = 1/(dG^(-1)/dx) = (1+u*x)/(1+(1+u)*x),
G(u,x)=exp[x*H(u,t)d/dt] t , evaluated at t=0. Also, dG(u,x)/dx = H(u,G(u,x)).
(End)
Contribution from Tom Copeland, Sep 16 2011: (Start)
f(u,z) and F(u,z) are expressible in terms of the incomplete gamma function Γ(v,p)(see Laplace Transforms for Power-law Functions at EqWorld):
With K(p,s) = p^(-s-1) exp(p) Γ(s+1,p),
f(u,z) = K(p,s)/(u*z) with p=(u+1)/(u*z) and s=1/z , and
F(u,z) = K(p,s)/(u*z) with p=(u+1)/(u^2*z) and s=1/(u^2*z).
(End)
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