From: Raymond Manzoni <raymman@club-internet.fr>
Subject: Re: Stirling Numbers and Partial Zeta Sums
Date: Fri, 27 Aug 1999 23:42:25 +0200
Newsgroups: sci.math

    Hi,

    I don't know if it's really what you want but following generating
function will give you (directly) the formula for s(m,k)

    exp(S_1*z/1-S_2*z^2/2+S_3*z^3/3+...) = 1
+S_1*z/1!+(S_n)*z^1/1!+(S_1^2-S_2)*z^2/2!+...

    This is easy to find using following formula :
    1/n! sum_{k>=0} s(n+1,k+1)*x^k = (1+x)(1+x/2)...(1+x/n) = exp(
sum_{j=1}^n log(1+x/j) )

    This is (implicitly) contained in following paper : S.Wilf "The
asymptotic behavior of the Stirling numbers of the first kind" page 2
available at ftp://ftp.cis.upenn.edu/pub/wilf/stirling.ps

    Have a look too at Knuth "The Art of computer programming I 1973"
page 91 for expressions like :
    sum_{1<=i<=j<=k<=n} x_i*x_j*x_k = 1/3!
[(sum_k(x_k))^3+3*sum_k(x_k)*sum_k(x_k^2)+2*sum_k(x_k^3)]

    (all signs became positive) corresponding to the generating function
:
        1/[(1-x_1*z)(1-x_2*z)...(1-x_n*z)]

    Hope it helped,

            Raymond Manzoni