%I #23 Mar 23 2020 02:43:05

%S 1,1,1,20,1,168,1680,1,440,924,55440,369600,1,910,10010,300300,

%T 1261260,33633600,168168000,1,1632,37128,48620,1113840,24504480,

%U 17153136,326726400,2058376320,34306272000,137225088000

%N Ordered set partitions of the set {1, 2, ..., 3*n} with all block sizes divisible by 3, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

%C T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 3. For instance 3*P(4, .) = [[12], [9, 3], [6, 6], [6, 3, 3], [3, 3, 3, 3]].

%e Triangle starts (note the subdivisions by ';' (A072233)):

%e [0] [1]

%e [1] [1]

%e [2] [1; 20]

%e [3] [1; 168; 1680]

%e [4] [1; 440, 924; 55440; 369600]

%e [5] [1; 910, 10010; 300300, 1261260; 33633600; 168168000]

%e [6] [1; 1632, 37128, 48620; 1113840, 24504480, 17153136; 326726400, 2058376320;

%e 34306272000; 137225088000]

%e .

%e T(4, 1) = 440 because [9, 3] is the integer partition 3*P(4, 1) in the canonical order and there are 220 set partitions which have the shape [9, 3]. Finally, since the order of the sets is taken into account, one gets 2!*220 = 440.

%o (Sage) # uses[GenOrdSetPart from A327022]

%o def A327023row(n): return GenOrdSetPart(3, n)

%o for n in (0..6): print(A327023row(n))

%Y Row sums: A243664, alternating row sums: A002115, main diagonal: A014606, central column A281479, by length: A278073.

%Y Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), this sequence (m=3), A327024 (m=4).

%Y Cf. A080577, A000041, A072233.

%K nonn,tabf

%O 0,4

%A _Peter Luschny_, Aug 27 2019