Bell #s Generalized: e^e^..^(e^x-1)

[Leroy Quet]

It seems like I have heard of this before. And the idea is obvious.
But the next-highest-order sequence above the standard Bell numbers is 
not in the on-line Encyclopedia of Integer Sequences, IF I have 
calculated the first few terms correctly (by hand).

It is known that, if B(k) is the k_th (standard) Bell number,

sum{k=0 to oo}  B(k) x^k /k!  =

exp(exp(x)-1).


But if we want, for a fixed m >= 2, {B(m,k)}, where

sum{k=0 to oo}  B(m,k) x^k /k!  =

exp(exp(...exp(exp(x)-1)..)),

with m 'exp's,

(therefore, B(2,k) is a standard Bell #)

then:

B(1,k) = 1 for all k >= 0 (violating the generating-function, due to 
B(1,0) = 1, and not 0);

B(m,0) = 1;

and:

B(m,j+1) = 


sum{k=0 to j} binomial(j,k)  B(m-1,k) B(m,j-k)



(If I have not erred)

By the way, {B(3,k)} to start:

1, 1, 2, 6, 23, 106,...

Am I even right about any of this? 
(If not, this would explain why {B(3,k)} is not in the EIS.)

 
Thanks,
Leroy Quet