A327023 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.

1, 1, 1, 20, 1, 168, 1680, 1, 440, 924, 55440, 369600, 1, 910, 10010, 300300, 1261260, 33633600, 168168000, 1, 1632, 37128, 48620, 1113840, 24504480, 17153136, 326726400, 2058376320, 34306272000, 137225088000

Offset: 0

Author

Peter Luschny, Aug 27 2019

Keywords

• nonn
• tabf

Comments

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]].

Examples

			Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1;   20]
[3] [1;  168;  1680]
[4] [1;  440,   924;  55440;  369600]
[5] [1;  910, 10010; 300300, 1261260; 33633600; 168168000]
[6] [1; 1632, 37128,  48620; 1113840, 24504480,  17153136; 326726400, 2058376320;
     34306272000; 137225088000]
.
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.
		

Crossrefs

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

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

Cf. A080577, A000041, A072233.

Programs

Sage

# uses[GenOrdSetPart from A327022]
def A327023row(n): return GenOrdSetPart(3, n)
for n in (0..6): print(A327023row(n))