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Number A(n,k) of set partitions of [k*n] such that within each block the numbers of elements from all residue classes modulo k are equal for k>0, A(n,0)=1; square array A(n,k), n>=0, k>=0, read by antidiagonals.
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%I #24 Jun 20 2021 21:04:39

%S 1,1,1,1,1,1,1,1,2,1,1,1,3,5,1,1,1,5,16,15,1,1,1,9,64,131,52,1,1,1,17,

%T 298,1613,1496,203,1,1,1,33,1540,25097,69026,22482,877,1,1,1,65,8506,

%U 461105,4383626,4566992,426833,4140,1,1,1,129,48844,9483041,350813126,1394519922,437665649,9934563,21147,1

%N Number A(n,k) of set partitions of [k*n] such that within each block the numbers of elements from all residue classes modulo k are equal for k>0, A(n,0)=1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%H Alois P. Heinz, <a href="/A275043/b275043.txt">Antidiagonals n = 0..60, flattened</a>

%H J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/bell.html">Extended Bell and Stirling Numbers From Hypergeometric Exponentiation</a>, J. Integer Seqs. Vol. 4 (2001), #01.1.4.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_of_a_set">Partition of a set</a>

%e A(2,2) = 3: 1234, 12|34, 14|23.

%e A(2,3) = 5: 123456, 123|456, 126|345, 135|246, 156|234.

%e A(2,4) = 9: 12345678, 1234|5678, 1238|4567, 1247|3568, 1278|3456, 1346|2578, 1368|2457, 1467|2358, 1678|2345.

%e A(3,2) = 16: 123456, 1234|56, 1236|45, 1245|36, 1256|34, 12|3456, 12|34|56, 12|36|45, 1346|25, 1456|23, 14|2356, 14|23|56, 16|2345, 16|23|45, 14|25|36, 16|25|34.

%e Square array A(n,k) begins:

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, 2, 3, 5, 9, 17, 33, ...

%e 1, 5, 16, 64, 298, 1540, 8506, ...

%e 1, 15, 131, 1613, 25097, 461105, 9483041, ...

%e 1, 52, 1496, 69026, 4383626, 350813126, 33056715626, ...

%e 1, 203, 22482, 4566992, 1394519922, 573843627152, 293327384637282, ...

%p A:= proc(n, k) option remember; `if`(k*n=0, 1, add(

%p binomial(n, j)^k*(n-j)*A(j, k), j=0..n-1)/n)

%p end:

%p seq(seq(A(n, d-n), n=0..d), d=0..12);

%t A[n_, k_] := A[n, k] = If[k*n == 0, 1, Sum[Binomial[n, j]^k*(n-j)*A[j, k], {j, 0, n-1}]/n]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* _Jean-François Alcover_, Jan 17 2017, translated from Maple *)

%Y Columns k=0-10 give: A000012, A000110, A023998, A061684, A061685, A061686, A061687, A061688, A275097, A275098, A275099.

%Y Rows n=0+1,2-5 give: A000012, A094373, A275100, A275101, A275102.

%Y Main diagonal gives A275044.

%Y Cf. A345400.

%K nonn,tabl

%O 0,9

%A _Alois P. Heinz_, Jul 14 2016