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<DIV>Franklin,</DIV>
<DIV> This is a good observation.
</DIV>
<DIV>In looking back at my emails with Benoit Cloitre (who came up with the
sum),</DIV>
<DIV>I noticed the same thing on 1/29/2003 (I should have updated A080093
then!).</DIV>
<DIV> </DIV>
<DIV>The e.g.f.s for the modified sequences are as follows: </DIV>
<DIV> </DIV>
<DIV>A080093 (after multiplying A080093(3*k+1) by 2, for k>=0): </DIV>
<DIV>E.g.f.: (exp(exp(x/2) - 1 - x/2) - 1)/2 </DIV>
<DIV> </DIV>
<DIV>A080094 (after multiplying A080094(3*k+1) by 2, for k>=0):</DIV>
<DIV>E.g.f.: (1 - exp(1 - x/2 - exp(x/2)))/2</DIV>
<DIV> </DIV>
<DIV>Thus, Benoits' sum becomes:</DIV>
<DIV> </DIV>
<DIV>Sum_{n>=0} Sum_{k>=0} k^n/(2*k+1)!)*x^n/n! = </DIV>
<DIV> (exp(exp(x/2) - 1 - x/2) - 1)/2 * exp(1) + </DIV>
<DIV> (1 - exp(1 - x/2 - exp(x/2)))/2 / exp(1)</DIV>
<DIV> </DIV>
<DIV>or, equivalently, we have the final form:</DIV>
<DIV> </DIV>
<DIV>Sum_{n>=0} Sum_{k>=0} k^n/(2*k+1)! *x^n/n!
= sinh(exp(x/2))/exp(x/2) </DIV>
<DIV> </DIV>
<DIV>Numerical check verifies this result - </DIV>
<DIV>and in fact the final formula is not too hard to prove. </DIV>
<DIV> </DIV>
<DIV>I agree with your propose updates.</DIV>
<DIV> </DIV>
<DIV>Paul</DIV>
<DIV> </DIV>
<DIV>On Wed, 10 Aug 2005 19:32:40 -0400 <A
href="mailto:franktaw@netscape.net">franktaw@netscape.net</A> (Franklin T.
Adams-Watters) writes:<BR>> I happened to look at new sequence A111579
today. The description of <BR>> this sequence needs to be clarified a
bit; the Q should be replaced <BR>> by Q(m,k), with the note that
Q(m,k)=(k-1)*m+1. The table is better <BR>> described as a square array
by antidiagonals, not as a triangle, and <BR>> then a(m,n) is the sum of the
terms in the nth row of the <BR>> generalized Stirling triangle using Q(m,k)
as the coefficients for <BR>> the kth column.<BR>> <BR>> Given this,
the zeroth row of the array (this is the zeroth column <BR>> from the
original description) is wrong. This row is given as all <BR>> ones,
which would correspond to a Q sequence of (0,0,0,...). <BR>> However,
the sequence should be (1,0,-1,-2,...) to be consistent <BR>> with the rest
of the array. A quick look at the columns of the <BR>> square array
(particularly the 2nd column, but every column is a <BR>> polynomial except
for the first row) shows that the values given do <BR>> not fit.<BR>>
<BR>> Now it starts to get interesting. When we build the sequence with
<BR>> (1,0,-1,-2,...), we get:<BR>>
1,2,3,3,2,3,5,-4,5,55,-212,201,2381,-15350,35183,145359,-1821438,...<BR>>
This sequence is not in the OEIS, but there is A080094:<BR>>
1,-3,3,-1,3,-5,-2,-5,55,106,201,-2381,-7675,-35183,145359,910719,...<BR>>
Offset by 1, this differs by a factor of (-1)^n, with an additional <BR>>
divisor of 2 every third term. Looking at A080093-A080095, we see <BR>>
that in fact A080094(n) is being divided by 2^(n+1), except that <BR>> every
third term is only divided by 2^n. And, of course, this is <BR>>
exactly the factor of 2 that distinguishes these terms.<BR>> <BR>>
"Restoring" this factor of 2 to A080093 gives us a match for <BR>>
A000296.<BR>> <BR>> So, can anyone prove either or both of these
correspondences (new <BR>> sequence to A080094, or A000296 to
A080093)?<BR>> <BR>> Assuming that such proofs can be produced, I think we
should make <BR>> the following changes:<BR>> * Add the sequence
above. Cross reference it with A080094.<BR>> * Cross reference A000296
with A080093.<BR>> * Add terms 1, -1, 1 respectively as index 0 terms to
<BR>> A080093-A080095. (Note that this matches the definition of these
<BR>> sequences: Sum 1/(2k+1)! = sinh(1) = e - 1/e.)<BR>> * Edit A111579
to remove (or fix) the first row, and fix the <BR>> description.<BR>> --
<BR>> Franklin T. Adams-Watters<BR>> 16 W. Michigan Ave.<BR>> Palatine,
IL 60067<BR>> 847-776-7645</DIV></BODY></HTML>