Exp-Generating-Function for x(exp(x(1+e^x)))

[Leroy Quet]

Posted to sci.math:
  
I am guessing that....

If a(m) =

sum{k=1 to m}  binomial(m,k) k^(m+1-k),

then we have a(m) is 

the m_th derivative of

x*exp(x(1+e^x))

at x = 0,

ie. sum{k=1 to oo} a(k) x^k /k!

= x(exp(x(1+e^x))).

This sequence {a(k)} begins: 1, 4, 18, 92, ...

Does this sequence have a simpler representation?

Now a(m)/m is simply sequence A080108 in the EIS.

But I am unsure if my particular sequence is in the EIS.

By the way, x(exp(x(1+e^x))) =

f(f(x)) if f(x) = x*exp(x),

ie. sum{k=1 to oo} a(k) (W(W(x))^k /k!  = x.

(W(x) is the Lambert function, of course.)

Is there an easy way to get a(n,m), where

sum{k=1 to oo} a(n,k) x^k /k! =

f(f(...f(x)..))   (with n f's)

if f(x) = x exp(x)?


Thanks,
Leroy Quet