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

[Frank Ruskey]

Take a look at A000258 for e^(e^(e^(x)-1)-1).
I think you need the extra -1.
-Frank

----------------------
Frank Ruskey                     e-mail: fruskey at cs.uvic.ca
Dept. of Computer Science        fax:    250-721-7292
University of Victoria           office: 250-721-7232
Victoria, B.C. V8W 3P6 CANADA    WWW: http://www.cs.uvic.ca/~fruskey

On Thu, 24 Jul 2003, Leroy Quet wrote:

> 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
>