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

0,4

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]].

Links

Table of n, a(n) for n=0..29.

Example

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.

Prog

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

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.

Sequence in context: A200092 A164812 A332258 * A280621 A223522 A040395

Adjacent sequences: A327020 A327021 A327022 * A327024 A327025 A327026

Keyword

nonn,tabf

Author

Peter Luschny, Aug 27 2019

Status

approved