A recurrence - Triangle

[Paul D. Hanna]

Brendan, Seqfans,
    One of the nice formulas that Max gave can obviously form a triangle:
> ss(n)=n!*sum(k=0,n,binomial(-x-2,k)/(n-k)!)
> For example, for n=0..10 we get the following values:
> ? vector(11,n,ss(n-1))
> %1 = [1, -x - 1, x^2 + 3*x + 3, -x^3 - 6*x^2 - 14*x - 11, x^4 + 
> 10*x^3
> + 41*x^2 + 76*x + 53, -x^5 - 15*x^4 - 95*x^3 - 305*x^2 - 489*x - 
> 309,
 
If row n of a triangle T(n,k) = coefficient of x^(n-k) in ss(n) as given
above, 
we obtain: 
1;
-1,-1;
3,3,1;
-11,-14,-6,-1;
53,76,41,10,1;
-309,-489,-305,-95,-15,-1;
2119,3623,2524,925,190,21,1;
-16687,-30414,-23121,-9569,-2345,-343,-28,-1; ...
which is not in the OEIS.
 
This triangle T can be expressed by 
   T = C*S*D 
where
C = Pascal's triangle 
S = A049444  ("Generalized Stirling number triangle of first kind")
D = diagonal matrix [1,-1,1,-1,...].
 
Just another alternate formula. 
FYI,
    Paul
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