

A343696


a(n) is the number of preference profiles in the stable marriage problem with n men and n women, such that the men's preference profiles form a Latin square.


6



1, 8, 2592, 191102976, 4013162496000000, 113241608573209804800000000, 5078594244241245901264634511360000000000, 759796697672599288560347581750936194390876487680000000000, 602809439070636186475532789128702956081602819845966698324215778508800000000000
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OFFSET

1,2


COMMENTS

Equivalently, these are the profiles where each woman is ranked differently by the n men.
Equivalently, for every rank i, there is exactly one woman who is ranked i by a given man.
The menproposing GaleShapley algorithm on such a set of preferences ends in one round, since every woman receives one proposal in the first round.
Due to symmetry, a(n) is the number of preference profiles in the stable marriage problem with n men and n women, such that the women’s preference profiles form a Latin square.


LINKS

Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, Sequences of the Stable Matching Problem, arXiv:2201.00645 [math.HO], 2021.


FORMULA



EXAMPLE

For n = 3, there are 3!^3 ways to set up the women's preference profiles and A002860(3) ways to set up the men's preference profiles, where A002860(3) = 12 (there are 12 different Latin squares of order 3). Thus a(3) = 3!^3 * A002860(3) = 216 * 12 = 2592.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



