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A350584 Triangle read by rows, T(n, k) = [x^k] ((2*x^3 - 3*x^2 - x + 1)/(1 - x)^(n + 2)), for n >= 1 and 0 <= k < n.

%I A350584 #11 Mar 27 2022 10:54:14
%S A350584 1,1,3,1,4,7,1,5,12,19,1,6,18,37,56,1,7,25,62,118,174,1,8,33,95,213,
%T A350584 387,561,1,9,42,137,350,737,1298,1859,1,10,52,189,539,1276,2574,4433,
%U A350584 6292,1,11,63,252,791,2067,4641,9074,15366,21658
%N A350584 Triangle read by rows, T(n, k) = [x^k] ((2*x^3 - 3*x^2 - x + 1)/(1 - x)^(n + 2)), for n >= 1 and 0 <= k < n.
%e A350584 Triangle starts:
%e A350584 [1] [1]
%e A350584 [2] [1,  3]
%e A350584 [3] [1,  4,  7]
%e A350584 [4] [1,  5, 12,  19]
%e A350584 [5] [1,  6, 18,  37,  56]
%e A350584 [6] [1,  7, 25,  62, 118,  174]
%e A350584 [7] [1,  8, 33,  95, 213,  387,  561]
%e A350584 [8] [1,  9, 42, 137, 350,  737, 1298, 1859]
%e A350584 [9] [1, 10, 52, 189, 539, 1276, 2574, 4433, 6292]
%p A350584 # Compare the analogue algorithm for the Bell triangle in A046937.
%p A350584 A350584Triangle := proc(len) local A, P, T, n; A := [2]; P := [1]; T := [[1]];
%p A350584 for n from 1 to len-1 do P := ListTools:-PartialSums([op(P), A[-1]]);
%p A350584 A := P; T := [op(T), P] od; T end:
%p A350584 A350584Triangle(10): ListTools:-Flatten(%);
%p A350584 # Alternative:
%p A350584 ogf := n -> (2*x^3 - 3*x^2 - x + 1)/(1 - x)^(n + 2):
%p A350584 ser := n -> series(ogf(n), x, n):
%p A350584 row := n -> seq(coeff(ser(n), x, k), k = 0..n-1):
%p A350584 seq(row(n), n = 1..10);
%Y A350584 A280891 (row sums), A135339 (alternating row sums), A005807 or A071716 (main diagonal).
%K A350584 nonn,tabl
%O A350584 1,3
%A A350584 _Peter Luschny_, Mar 27 2022