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%I #13 Jul 09 2023 12:08:23
%S 1,1,1,2,1,1,6,2,2,1,1,24,6,6,2,2,1,1,120,24,24,24,6,6,6,2,2,1,1,720,
%T 120,120,120,24,24,24,24,6,6,6,2,2,1,1,5040,720,720,720,720,120,120,
%U 120,120,120,24,24,24,24,24,6,6,6,2,2,1,1,40320
%N Irregular table T(n,k) = A178886(n,k)/A048996(n,k) read by rows.
%C Row n has A000041(n) terms.
%C Consider the five partitions of the number 4:
%C 4 3+1 2+2 2+1+1 and 1+1+1+1
%C rewriting as 4000 3100 2200 2110 and 1111
%C then a(n) counts the ways that the zeros can be permuted:
%C 6,2,2,1,1
%C agreeing with the factorial of the difference between A036042 and A036043.
%F T(n,k) = ( A036042(n,k) - A036043(n,k))!.
%F T(n,k) = n!/A178888(n,k). - _R. J. Mathar_, Mar 03 2011
%e Row four of A178886 begins: 6 4 2 3 1
%e Row four of A048996 begins: 1 2 1 3 1
%e so,
%e Row four of A179972 begins: 6 2 2 1 1
%e Triangle T(n,k) begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 6, 2, 2, 1, 1;
%e 24, 6, 6, 2, 2, 1, 1;
%e 120, 24, 24, 24, 6, 6, 6, 2, 2, 1, 1;
%e ...
%Y Cf. A178886, A048996, A036042, A036043, A179973 (row sums).
%K nonn,tabf
%O 1,4
%A _Alford Arnold_, Aug 04 2010
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