Eulerians as Weighted Binomial Coefficients

There's a nice relation between Eulerian numbers and the generalized
Bernoulli numbers.  Let W(m,n) be a weighted average of the nth powers 
of the first m natural numbers, using the Eulerians as the weights.  
For example

           (1)^5 + 26(2)^5 + 66(3)^5 + 26(4)^5 + (5)^5         777
 W(6,5) =  ---------------------------------------------  =   -----
                               120                              2

For m>n it turns out that W(m,n) equals B(-m,n), the generalized 
Bernoulli number, which can be defined as the coefficients in the 
expansion 
                                    inf   B(-m,n)
            ( x/(e^x - 1) )^-m  =  SUM   --------- x^n
                                    n=0      n!

Beginning at the "diagonal" where m=n, the values of W and B begin to
differ, although they differ only half the time on the diagonal.
Specifically, the differences B(-m,m) - W(m,m) for m=0,1,2,3... along 
the diagonal are

 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, ...

which of course are just the ordinary Bernoulli numbers.