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%I #12 May 03 2026 23:06:33
%S 1,0,1,0,3,1,0,3,6,1,0,3,24,9,1,0,3,60,63,12,1,0,3,132,333,120,15,1,0,
%T 3,276,1359,984,195,18,1,0,3,564,4869,6600,2175,288,21,1,0,3,1140,
%U 16263,37272,20715,4068,399,24,1
%N Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} Stirling2(n, j)*3^j*(k)_j.
%C Base 3 polynomial collapse array for the iterated inverse Pascal operator.
%C For base 2, the analogous one-step array is A394444.
%F Let D be the inverse Pascal operator: [D(f)]_m = Sum_{i=0..m} (-1)^(m-i)*binomial(m, i)*f(i). Then A(n, k) = [D^2(i^n*3^i)]_k = Sum_{j=0..n} Stirling2(n, j) *3^j*(k)_j.
%F Seen as a triangle, T(n, k) = A(n - k, k).
%F For fixed k, the e.g.f. of A(n, k) in n is Sum_{n>=0} A(n, k)*u^n/n! = (1 + 3*(exp(u) - 1))^k.
%e k | 0 1 2 3 4 5
%e n |------------------------------------------------
%e 0 | 1 1 1 1 1 1
%e 1 | 0 3 6 9 12 15
%e 2 | 0 3 24 63 120 195
%e 3 | 0 3 60 333 984 2175
%e 4 | 0 3 132 1359 6600 20715
%e 5 | 0 3 276 4869 37272 169575
%e A(2, 2) = Stirling2(2, 1)*3*(2)_1 + Stirling2(2, 2)*9*(2)_2 = 24.
%Y Cf. A394444, A008277, A152751, A182464.
%K nonn,tabl
%O 0,5
%A _Dalton Heilig_, Apr 28 2026