

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.


3



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
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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 kth integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0based 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



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



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



