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%I #47 Sep 07 2020 09:35:02
%S 0,0,1,4,22,140,1020,8400,77280,786240,8769600,106444800,1397088000,
%T 19718899200,297859161600,4794806016000,81947593728000,
%U 1482030950400000,28277150533632000,567677135241216000,11961768206868480000,263969867887165440000
%N Total number of inversions in all permutations of order n consisting of a single cycle.
%C The formula trivially follows from the observation that every pair of elements i<j forms an inversion in exactly (binomial(n,2)-n+j-i)*(n-3)! single-cycle permutations. - _Max Alekseyev_, Jan 05 2018
%C a(n) is the number of ways to partition a (n+1)X(n+1) square, with the upper left hand corner missing, into ribbons of size n, see Alexandersson, Jordan. - _Per W. Alexandersson_, Jun 02 2020
%H Alois P. Heinz, <a href="/A227404/b227404.txt">Table of n, a(n) for n = 0..449</a>
%H Per Alexandersson, Linus Jordan, <a href="https://arxiv.org/abs/1805.09778">Enumeration of border-strip decompositions</a>, arXiv:1805.09778 [math.CO], 2018.
%H Per Alexandersson, Linus Jordan, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Alexandersson/alex4.html">Enumeration of border-strip decompositions</a>, Journal of Integer Sequences, Vol. 22 (2019), Article 19.4.5.
%F For n>2, a(n) = n! * (3*n-1)/12. - _Vaclav Kotesovec_, Feb 14 2014
%e a(3) = 4 because the cyclic 3-permutations: (1,2,3), (1,3,2) written in one line (sequence) notation: {2,3,1}, {3,1,2} have 2 + 2 = 4 inversions.
%t Table[Total[Map[Inversions,Map[FromCycles,Map[List, Map[Prepend[#,n]&, Permutations[n-1]]]]]],{n,1,8}]
%Y Cf. A001809, A211606, A216239.
%K nonn
%O 0,4
%A _Geoffrey Critzer_, Sep 21 2013
%E a(13)-a(15) from _Alois P. Heinz_, Sep 26 2013
%E Terms a(16) and beyond from _Max Alekseyev_, Jan 05 2018
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