%I
%S 1,1,1,0,2,1,1,0,3,1,1,4,0,4,1,2,5,10,0,5,1,9,12,15,20,0,6,1,
%T 9,63,42,35,35,0,7,1,50,72,252,112,70,56,0,8,1,267,450,324,
%U 756,252,126,84,0,9,1,413,2670,2250,1080,1890,504,210,120
%N Triangle T(n,k) read by rows: inverse of the matrix PE = exp(P)/exp(1) given in A011971.
%C The structure of the triangle is A[r,c] = A000587(1+(rc))*binomial(r1,c1) where row index r and columnindex c start at 1.
%C Coefficients of polynomials defined recursively: P(0,x)=1, P(n+1,x)=x*P(n,x)P(n,x+1). All generated polynomials appear to be irreducible. Polynomials evaluated at x=c give sequences with e.g.f. exp(1cxexp(x)).
%H S. de Wannemacker, T. Laffey and R. Osburn, <a href="http://arXiv.org/abs/math.NT/0608085">On a conjecture of Wilf</a>
%F Let P be the lowertriangular Pascalmatrix, PE = exp(PI) a matrix exponential in exact integer arithmetic (or PE = lim exp(P)/exp(1) as limit of the exponential) then A= PE^1 and a(n) = A[n, read sequentially].  Gottfried Helms, Apr 08 2007
%e Triangle starts:
%e 1,
%e 1,1,
%e 0,2,1,
%e 1,0,3,1,
%e 1,4,0,4,1,
%e 2,5,10,0,5,1,
%e 9,12,15,20,0,6,1,
%e 9,63,42,35,35,0,7,1,
%Y First column is A000587 (Uppuluri Carpenter numbers) which is also the negative of the row sums (=P(n, 1)). Polynomials evaluated at 2 are A074051, at 1 A109747.
%K easy,tabl,sign
%O 0,5
%A _Gottfried Helms_, Apr 08 2007
%E Edited by _Ralf Stephan_, May 12 2007
