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

[Leroy Quet]

I wrote:
>....
>
>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)
>
>

Those who have replied about my errors are correct.


The recursion-sum should have been:


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

{with a 'k+1' replacing a 'k' as an index} 

(unless I am again incorrect)


Then, as noted by others, we should have the generating function with -1 
added at every level of exponentation.
 

By the way, the original erroneous recursion-sum gives the sequences 
where {B(r,k)} has a EGF of

b_r(x) = exp(integral{0 to x} b_{r-1}(y) dy)

(I think),

where b_1(x) = exp(x).

(b_2(x) is still e^(e^x-1), in any case.) 

Now THIS original/erroneous sequence, at least {B(3,k)}, is not yet in 
the EIS.
:)

Thanks,
Leroy
Quet

>
>(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.)