%I #11 Jan 19 2020 06:27:45
%S 1,-1,1,1,-1,1,-1,1,1,1,1,-1,1,7,1,-1,1,1,21,21,1,1,-1,1,51,161,51,1,
%T -1,1,1,113,813,813,113,1,1,-1,1,239,3361,7631,3361,239,1,-1,1,1,493,
%U 12421,53833,53833,12421,493,1,1,-1,1,1003,42865,320107,607009,320107,42865
%N Triangle read by rows: expansion of (1 - x)/(exp(t)*(1 - x*exp(t*(1 - x))))
%C This sequence was derived from the Eulerian number umbral calculus expansion and A046802 by taking the exp(t) term and inverting it.
%C What is interesting here is the '1,-1' terms that appear.
%C I had thought I would get "1,5,1" not "1,7,1" from this function.
%C An OEIS search came up with A046739 which has the same internal symmetric number structure.
%C Inverse binomial transform of Eulerian numbers A123125. [Paul Barry, May 10 2011]
%F E.g.f. sum(T(n,k) t^n/n! x^k) = p(x,t) = (1 - x)/(exp(t)*(1 - x*exp(t*(1 - x))))
%F T(n,k)=sum{j=0..n, (-1)^(n-j)*C(n,j)*A123125(j,k)}. [_Paul Barry_, May 10 2011]
%e {1},
%e {-1, 1},
%e {1, -1, 1},
%e {-1, 1, 1, 1},
%e {1, -1, 1, 7, 1},
%e {-1, 1, 1, 21, 21, 1},
%e {1, -1, 1, 51, 161, 51, 1},
%e {-1, 1, 1, 113, 813, 813, 113, 1},
%e {1, -1, 1, 239, 3361, 7631, 3361, 239, 1},
%e {-1, 1, 1, 493, 12421, 53833, 53833, 12421, 493, 1},
%e {1, -1, 1, 1003, 42865, 320107, 607009, 320107, 42865, 1003, 1}
%t p[t_] = (1 - x)/(Exp[t]*(1 - x*Exp[t*(1 - x)]))
%t a = Table[ CoefficientList[FullSimplify[ExpandAll[n!*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]]], x], {n, 0, 10}];
%t Flatten[a]
%Y Cf. A046802, A046739, A000166 (row sums), A123125.
%K sign,tabl
%O 0,14
%A _Roger L. Bagula_, Nov 25 2009