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%I #8 Sep 01 2022 17:29:13
%S 1,0,1,0,3,2,0,13,18,6,0,73,158,108,24,0,501,1510,1590,720,120,0,4051,
%T 15962,23040,15960,5400,720,0,37633,186270,345786,325920,168000,45360,
%U 5040,0,394353,2385182,5469492,6579384,4594800,1884960,423360,40320
%N Triangle read by rows. T(n, k) = k! * Sum_{j=k..n} Lah(n, j) * Stirling2(j, k), where Lah(n, k) = A271703(n, k).
%C The same construction with Stirling1 in place of Stirling2 gives A225479, the ordered Stirling cycle numbers.
%e Triangle T(n, k) begins:
%e [0] 1;
%e [1] 0, 1;
%e [2] 0, 3, 2;
%e [3] 0, 13, 18, 6;
%e [4] 0, 73, 158, 108, 24;
%e [5] 0, 501, 1510, 1590, 720, 120;
%e [6] 0, 4051, 15962, 23040, 15960, 5400, 720;
%e [7] 0, 37633, 186270, 345786, 325920, 168000, 45360, 5040;
%e [8] 0, 394353, 2385182, 5469492, 6579384, 4594800, 1884960, 423360, 40320;
%p L := (n, k) -> `if`(n = k, 1, binomial(n-1, k-1) * n! / k!):
%p T := (n, k) -> k! * add(L(n, j) * Stirling2(j, k), j = k..n):
%p seq(seq(T(n, k), k = 0..n), n = 0..9);
%t T[n_, k_] := k! * Sum[Binomial[n, j] * FactorialPower[n - 1, n - j] * StirlingS2[j, k], {j, k, n}]; Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Sep 01 2022 *)
%Y Cf. A271703, A048993, A225479, A000262 (column 1), A052838 (column 2), A084358 (row sums).
%K nonn,tabl
%O 0,5
%A _Peter Luschny_, Sep 01 2022
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