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%I #20 Mar 23 2020 02:43:11
%S 1,1,1,70,1,990,34650,1,3640,12870,2702700,63063000,1,9690,251940,
%T 26453700,187065450,17459442000,305540235000,1,21252,1470942,2704156,
%U 154448910,8031343320,9465511770,374796021600,3975514943400,231905038365000,3246670537110000
%N Ordered set partitions of the set {1, 2, ..., 4*n} with all block sizes divisible by 4, 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 = 4. For instance 4*P(4, .) = [[16], [12, 4], [8, 8], [8, 4, 4], [4, 4, 4, 4]].
%e Triangle starts (note the subdivisions by ';' (A072233)):
%e [0] [1]
%e [1] [1]
%e [2] [1; 70]
%e [3] [1; 990; 34650]
%e [4] [1; 3640, 12870; 2702700; 63063000]
%e [5] [1; 9690, 251940; 26453700, 187065450; 17459442000; 305540235000]
%e [6] [1; 21252, 1470942, 2704156; 154448910, 8031343320, 9465511770;
%e 374796021600, 3975514943400; 231905038365000; 3246670537110000]
%e .
%e T(4, 1) = 3640 because [12, 4] is the integer partition 4*P(4, 1) in the canonical order and there are 1820 set partitions which have the shape [12, 4]. Finally, since the order of the sets is taken into account, one gets 2!*1820 = 3640.
%o (Sage) # uses[GenOrdSetPart from A327022]
%o def A327024row(n): return GenOrdSetPart(4, n)
%o for n in (0..6): print(A327024row(n))
%Y Row sums: A243665, alternating row sums: A211212, main diagonal: A014608, central column: A281480, by length: A278074.
%Y Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), A327023 (m=3), this sequence (m=4).
%Y Cf. A080577, A000041, A072233.
%K nonn,tabf
%O 0,4
%A _Peter Luschny_, Aug 27 2019
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