eulerian numbers for negative argmuents

[Mitch Harris]

>  it is a time ago, that I fiddled with that numbers. At
> http://mathpages.com/home/kmath012/kmath012.htm or
>    ...kmath464... there is also some stuff.

nice!

> From one of my q&d -tables I just copy&paste a list, which
> shows a relatively simple pattern how to compute the
> Euler-numbers for an arbitrary row just by row and col
> index. I'm too lazy to make that formula for you... it's
> way off my head for a long time.
> 
> 
>         Z      e(z,0)              e(z,1)               e(z,2)                   e(z,3)                              e(z,4)
>------------------------------------------------------------------------
------------------------------------------------------------------------
--------------------------------
>          -4 |  1*1^-4  |    1*2^-4 + 3*1^-4   |    1*3^-4 + 3*2^-4 +
6*1^-4    |      .4^-4 + 3*3^-4 +  6*2^-4 + 10*1^-4   |    1*5^-4 +
3*4^-4 +  6*3^-4 + 10*2^-4 + 15*1^-4
>          -3 |  1*1^-3  |    1*2^-3 + 2*1^-3   |    1*3^-3 + 2*2^-3 +
3*1^-3    |      1.4^-3 + 2*3^-3 +  3*2^-3 +  4*1^-3   |    1*5^-3 +
2*4^-3 +  3*3^-3 +  4*2^-3 +  5*1^-3
...

The formula for that table is:

   e(n, m) = sum_{k=0}^m b(n+1,k) (m+1-k)^n (-1)^k

(with shifting of n back to the usual function)

This can be read off the table almost directly, but I'm sure it is in 
Graham, Knuth, Patashnik, Concrete Mathematics. I hadn't yet considered 
it. That's a nice formula allowing negative n, which (fortunately!) 
computes the same values as I got.

Thanks for the lead. If I can't get a (bivariate) recurrence out of it, 
at least there's a summation formula (which is opposite of what one 
usually wants).

Mitch