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