Generalized Bell Numbers

[Paul D. Hanna]

Franklin,
       This is a good observation.  
In looking back at my emails with Benoit Cloitre (who came up with the
sum),
I noticed the same thing on 1/29/2003 (I should have updated A080093
then!).
 
The e.g.f.s for the modified sequences are as follows: 
 
A080093 (after multiplying A080093(3*k+1) by 2, for k>=0):  
E.g.f.:  (exp(exp(x/2) - 1 - x/2) - 1)/2 
 
A080094 (after multiplying A080094(3*k+1) by 2, for k>=0):
E.g.f.:  (1 - exp(1 - x/2 - exp(x/2)))/2
 
Thus, Benoits' sum becomes:
 
Sum_{n>=0} Sum_{k>=0} k^n/(2*k+1)!)*x^n/n!  = 
    (exp(exp(x/2) - 1 - x/2) - 1)/2 * exp(1) + 
    (1 - exp(1 - x/2 - exp(x/2)))/2 / exp(1)
 
or, equivalently, we have the final form:
 
Sum_{n>=0} Sum_{k>=0} k^n/(2*k+1)! *x^n/n!  =  sinh(exp(x/2))/exp(x/2) 
   
Numerical check verifies this result - 
and in fact the final formula is not too hard to prove. 
 
I agree with your propose updates.
 
Paul

On Wed, 10 Aug 2005 19:32:40 -0400 franktaw at netscape.net (Franklin T.
Adams-Watters) writes:
> I happened to look at new sequence A111579 today.  The description of 
> this sequence needs to be clarified a bit; the Q should be replaced 
> by Q(m,k), with the note that Q(m,k)=(k-1)*m+1.  The table is better 
> described as a square array by antidiagonals, not as a triangle, and 
> then a(m,n) is the sum of the terms in the nth row of the 
> generalized Stirling triangle using Q(m,k) as the coefficients for 
> the kth column.
> 
> Given this, the zeroth row of the array (this is the zeroth column 
> from the original description) is wrong.  This row is given as all 
> ones, which would correspond to a Q sequence of (0,0,0,...).  
> However, the sequence should be (1,0,-1,-2,...) to be consistent 
> with the rest of the array.  A quick look at the columns of the 
> square array (particularly the 2nd column, but every column is a 
> polynomial except for the first row) shows that the values given do 
> not fit.
> 
> Now it starts to get interesting.  When we build the sequence with 
> (1,0,-1,-2,...), we get:
> 1,2,3,3,2,3,5,-4,5,55,-212,201,2381,-15350,35183,145359,-1821438,...
> This sequence is not in the OEIS, but there is A080094:
> 1,-3,3,-1,3,-5,-2,-5,55,106,201,-2381,-7675,-35183,145359,910719,...
> Offset by 1, this differs by a factor of (-1)^n, with an additional 
> divisor of 2 every third term.  Looking at A080093-A080095, we see 
> that in fact A080094(n) is being divided by 2^(n+1), except that 
> every third term is only divided by 2^n.  And, of course, this is 
> exactly the factor of 2 that distinguishes these terms.
> 
> "Restoring" this factor of 2 to A080093 gives us a match for 
> A000296.
> 
> So, can anyone prove either or both of these correspondences (new 
> sequence to A080094, or A000296 to A080093)?
> 
> Assuming that such proofs can be produced, I think we should make 
> the following changes:
> * Add the sequence above.  Cross reference it with A080094.
> * Cross reference A000296 with A080093.
> * Add terms 1, -1, 1 respectively as index 0 terms to 
> A080093-A080095.  (Note that this matches the definition of these 
> sequences: Sum 1/(2k+1)! = sinh(1) = e - 1/e.)
> * Edit A111579 to remove (or fix) the first row, and fix the 
> description.
> -- 
> Franklin T. Adams-Watters
> 16 W. Michigan Ave.
> Palatine, IL 60067
> 847-776-7645
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