

A227404


Total number of inversions in all permutations of order n consisting of a single cycle.


3



0, 0, 1, 4, 22, 140, 1020, 8400, 77280, 786240, 8769600, 106444800, 1397088000, 19718899200, 297859161600, 4794806016000, 81947593728000, 1482030950400000, 28277150533632000, 567677135241216000, 11961768206868480000, 263969867887165440000
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OFFSET

0,4


COMMENTS

The formula trivially follows from the observation that every pair of elements i<j forms an inversion in exactly (binomial(n,2)n+ji)*(n3)! singlecycle permutations.  Max Alekseyev, Jan 05 2018
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


LINKS



FORMULA



EXAMPLE

a(3) = 4 because the cyclic 3permutations: (1,2,3), (1,3,2) written in one line (sequence) notation: {2,3,1}, {3,1,2} have 2 + 2 = 4 inversions.


MATHEMATICA

Table[Total[Map[Inversions, Map[FromCycles, Map[List, Map[Prepend[#, n]&, Permutations[n1]]]]]], {n, 1, 8}]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



