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%I #18 Jul 10 2023 07:42:01
%S 1,-1,2,2,-6,5,-6,22,-30,15,24,-100,175,-150,52,-120,548,-1125,1275,
%T -780,203,720,-3528,8120,-11025,9100,-4263,877,-5040,26136,-65660,
%U 101535,-101920,65366,-24556,4140,40320,-219168,590620,-1009260,1167348,-920808,478842,-149040,21147
%N Exponential transform of Stirling1 triangle A008275.
%C |a(n,k)| = number of sets of permutations of {1,...,n} with k total cycles.
%C From _David Callan_, Sep 20 2007: (Start)
%C |a(n,k)| = Stirling1(n, k) * Bell(k) counts the above sets of permutations. To see this, recall that Stirling1(n, k) is the number of permutations of [n]={1,...,n} with k cycles and Bell(k) is the number of set partitions of [k].
%C Given such a permutation and set partition, write the permutation in standard cycle form (smallest entry first in each cycle and first entries decreasing left to right). For example, with n=15 and k=6, {{10}, {6, 11}, {5, 7, 15}, {3, 13, 12, 8}, {2, 14, 9}, {1, 4}} is in this standard cycle form.
%C Then combine cycles as specified by the partition to form a set of lists. For example, the partition 156-24-3 would yield {{10, 2, 14, 9, 1, 4}, {6, 11, 3, 13, 12, 8}, {5, 7, 15}}. The original first entries are now the record left-to-right lows.
%C Finally, apply to each list the well known transformation that sends # record lows to # cycles. The example yields {{4, 14, 1, 2, 10, 9}, {13, 11, 3, 6, 8, 12}, {7, 15, 5}}. This is a bijection to sets of lists (i.e. permutations) with a total of k cycles, as required. (End)
%F E.g.f.: exp((1+x)^y-1).
%F a(n,k) = Stirling1(n,k) * Bell(k). - _Vladeta Jovovic_, Feb 01 2003
%e Triangle begins:
%e 1;
%e -1, 2;
%e 2, -6, 5;
%e -6, 22, -30, 15;
%e 24, -100, 175, -150, 52;
%e ...
%e |a(3,2)| = 6 because (12)(3), (12)|(3), (13)(2), (13)|(2), (23)(1), (23)|(1).
%Y Row sums of |a(n,k)| give A000262.
%Y Cf. A000110, A008275, A008297, A055925, A104150.
%K sign,tabl
%O 1,3
%A _Wouter Meeussen_, _Christian G. Bower_, Jul 06 2000
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